LIBRARY OF CONGRESS, 



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UNITED STATES OF AMERICA. 



ELEMENTS 



OF 



PLANE AND SPHERICAL 

TRIGONOMETRY 

A TEXT-BOOK FOR COLLEGES AND SCHOOLS 

BY 

EDWIN S. CRAWLEY, Ph.D. 

ASSISTANT PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF PENNSYLVANIA 






SECOND EDITION, REVISED AND ENLARGED 



SEP 191896 



, 






>4 V> 

UNIVERSITY OF PENNSYLVANIA 

PHILADELPHIA 

1896 



Copyright, 1890, by Edwin S. Crawley 



Copyright, 1896, by Edwin S. Crawley 



Electrotyped and Printed by J. B. Uppincott Company Philadelphia, U.S.A. 

M 



PREFACE. 



Several years' experience in the class-room with the first 
edition of this book has shown the desirability of a number 
of alterations and additions. The most important of these 
are : (1) The adoption at the beginning of ratio definitions 
of the trigonometric functions that are applicable to angles 
of any magnitude, instead of a twofold set of definitions, 
first for angles less than 90°, and afterwards for angles in 
general. (2) The addition of a large number of exercises, 
worked in the text either wholly or in part, to illustrate the 
best methods of trigonometric reduction and analysis. (3) 
A large increase in the number and variety of the examples. 
(4) The addition of theorems on the escribed circles and 
Brocard's points. (5) A new chapter devoted to De Moivre's 
theorem and some of the deductions therefrom, and an intro- 
duction to hyperbolic functions. 

The book is intended to give, clearly and concisely, as 
much of the subject of trigonometry as the time usually 
allotted to it in our colleges will permit. A briefer course 
for those having less time at their disposal may be had by 
omitting some of the more advanced portions, without im- 
pairing the continuity of the subject. The parts that may 
be omitted for this purpose are indicated in the note below. 

The Appendix, containing a list of those formulae most 
frequently employed in subsequent mathematical study, will 
be found useful for reference. 

3 



4 PREFACE. 

Answers are given, at the end of the book, to the greater 
number of the examples. The answers are not given in 
those eases where it would detract from the value of the 
examples, or where it is thought best that the student should 
verify the results himself. 

The author desires to express his cordial thanks to his 
colleagues, Prof. Charles L. Doolittle, Dr. George H. Hallett, 
and Mr. Henry B. Evans, and especially to Dr. I. J. Schwatt, 
for valuable suggestions and kind assistance in the prepara- 
tion of this book. 

EDWIN S. CEAWLEY. 

University of Pennsylvania, 
August, 1896. 

Note. — Those wishing a shorter course than that given by 
the complete text may omit any or all of the following sec- 
tions : 8, 34, 40 or 41, 42, 44, 50, 51, 60, 65-67, 68-71, 74-86, 
106, 107, 114, 116, 118, 120, 124-126. 



TABLE OF CONTENTS, 



PART I. 



PLANE TEIGONOMETEY. 

CHAP. PAGE 

I. — The Measurement of Angles and Arcs . 7 

II. — The Trigonometric Functions 15 

III. — General Formulae 46 

IV.— Solution of Eight Triangles 61 

V. — Formulae for the Solution of Oblique Triangles 72 

VI.— Solution of Oblique Triangles 82 

VII. — Miscellaneous Problems and Trigonometric Equations . . 97 
VIII. — De Moivre's Theorem, Trigonometric Series, Hyperbolic 

Functions . . . . . 109 

PAET II. 

SPHEKICAL TKIGONOMETEY. 

Introduction 120 

IX.— General Formulae 122 

X.— Solution of Spherical Eight Triangles 140 

XI. — Solution of Spherical Oblique Triangles ......... 150 

XII. — Area, Circumscribed and Inscribed Circles . „ 165 



APPENDIX. 

List of Formula? 171 

Answers to Examples 173 

1* 5 



Elements of Trigonometry. 



RAR.T I. 

PLANE TRIGONOMETRY 



CHAPTEE I. 



THE MEASUREMENT OF ANGLES. 



1. Plane Trigonometry is, primarily, the investigation 
of the relations between the six parts (the three sides and the 
three angles) of a plane triangle, so that when three of these 
parts are known the other three may be computed. This 
subject is treated in geometry, but the methods there em- 
ployed, being purely graphical, cannot be used to obtain 
accurate numerical results. 

Plane Trigonometry investigates also the properties and 
relations of angular quantities in general, without reference to 
triangles. 

2. Angles. The conception of a plane angle, with which 
the student has become familiar in geometry, is not sufficiently 
general for the purposes of trigonometry. We need here a 
conception that will admit of angles up to any size we please, 

7 



8 



PLANE TRIGONOMETRY. 



FlG. 1. 




without limit. We can form such a conception of angles, or 
angular quantity, in the following manner : 

Let CA, Fig. 1, be a line whose 
position is fixed. Let another 
line, which we shall call CP, 
be conceived as coincident with 
CA. This line CP is a movable 
line, capable of revolving about 
the point C as the hand of a 
watch revolves about the centre 
of the dial. Now, let CP revolve 
in the direction of the arrow 
about C from its original posi- 
tion of coincidence with CA into 

a position CP r In this way an angle, ACP V is generated. 
As the line continues to revolve, the angle generated by it 
increases in magnitude. When it reaches the position CB 
perpendicular to CA, it forms with CA a right angle. The 
right angle is divided into 90 equal parts, called degrees (°), 
the degree is divided into 60 equal parts, called minutes ('), 
and the minute is divided into 60 equal parts, called seconds 
("). When still more minute divisions of the angle are re- 
quired, decimals of a second are used. Thus, we write 72° 
19' 38".74. 

If, now, we imagine the movable line CP to depart from its 
present position CB and to move onward toward CA V pass- 
ing through the position CP 2 , the angle it forms with CA will 
continually increase until, when it has reached CA V it will 
have described two right angles and its angle with CA will be 



180°. If a^ain the revolution be continued, 



shall have, 



when CP reaches the position CP V an angle, ACP V greater 
than two right angles. In geometry, under the same circum- 
stances, we should say that the angle between CA and CP 3 
was the angle ACP V less than 180° ; but it can easily be seen 
that to follow logically the conception with which we have 
started we must here take the angle greater than 180° as the 
anffle between CA and CP . The student will be better able 



THE MEASUREMENT OF ANGLES. 9 

to appreciate the advantages of this conception when he has 
seen some of its applications. 

Continuing the same process, we see that CB 1 makes with 
CA an angle of 270° ; that CP i makes with CA an angle 
greater than 270° ; and, finally, when the movable line CP 
has returned to its original position CA, it is said to make an 
angle of 360° with CA. 

3. Angles > 360°. If we follow our conception further it 
will lead us to important and still more general results. Let 
us suppose that after completing one revolution the movable 
line CP revolves 120° farther 

into the position OP, Fig. 2. In FlG> 2 * 
a geometric sense it now makes 
with CA the same angle that it 
would make if, starting from its 
original position CA, it had re- 
volved through only 120°. But A 

in the broader sense that we 

have adopted here it makes with 

CA an angle of 360° -f 120° = 

480°. In the same way we can 

conceive that the line CP has l 

revolved twice through 360° and then to its position on the 

figure. In that case the angle ACP= 2 X 360° + 120° = 

840°. Or, to be quite general, we say that the angle A CP is 

n X 360° -f- 120°, where n is any positive integer or zero. 

What is here said of 120° applies equally well to all angles. 
AYe may therefore express any angle in as many ways as 
we please by simply adding multiples of 360° to its smallest 
numerical expression. Thus, 0, 360 -f- 0, 2 X 360 -f- 0, n X 
360° -j- 0, all mean the same angle in the geometric sense ; but 
we shall meet with instances in which a distinction will have 
to be made between them. 

4. Definitions. It is customary to designate the four 
parts into which the plane is divided, Figs. 1 or 2, by the lines 
AA V BB 1 as the four quadrants; of which A CB is called the 



10 PLANE TRIGONOMETRY. 

first; BCA V the second; A 1 CB V the third; and B X CA, the 
fourth. 

Hence an angle between 0° and 90°, between 360° and 360° 
-f 90°, between 720° and 720° -f 90°, etc., is said to be an 
angle of the first quadrant: e.g., 72°, 415°, 786°, 1102°. An 
angle between 90° and 180°, between 360° -j- 90° and 360° -f 
180°, between 720° + 90° and 720° + 180°, etc., is said to be 
an angle of the second quadrant : e.g., 136°, 863°. An angle 
between 180° and 270°, between 360° -f 180° and 360° -f 270°, 
etc., is said to be an angle of the third quadrant: e.g., 216°, 
560°, 937°. An angle between 270° and 360°, between 360° 
+ 270° and 360° -f 360°, etc., is said to be in the fourth quad- 
rant : e.g., 315°, 1062°. 

5. Negative Angles. We have so far considered the 
movable line as revolving in only one direction, that contrary 
to the motion of the hands of a watch.* We can equally well 
regard it as performing its revolution in the opposite direc- 
tion ; that is, in the same direction as the hands of a watch.f 
We distinguish an angle conceived as generated in this way 
by a negative sign. 

In Fig. 3 the angle AGP, Fig. 3. 

which is supposed to have been b 

generated by CP moving from 
the position CA in the direction 
of the arrow, is a negative angle, 
and is expressed numerically 
as — 6. All that has been said * 
above of positive angles applies > 

to negative ones also. Thus, / 

A CP may be written —6, — 360° / 

— 6, — 2 X 360° — 6, — n X A 

360° — 0. 

To tell in what quadrant a given negative angle is found, 
we must remember that the first quadrant in the negative 

* Called "counter-clockwise." 
f Called " clockwise." 



THE MEASUREMENT OF ANGLES. 11 

direction is the fourth quadrant, the second in the negative 
direction is the third, and so on. Thus, — 123° is in the third 
quadrant, — 287° is in the first quadrant, — 561° is in the 
second, — 783° is in the fourth, etc. 

6. Measurement of Arcs. Since by elementary geom- 
etry the angle at the centre of the circle and its intercepted 
arc have the same measure, all that is said above of angles 
applies with equal force to arcs of circles. 

7. Definitions. Two angles or arcs whose sum is 90° are 
called complements of each other. 

Two angles or arcs whose sum is 180° are called supplements 
of each other. 

8. Centesimal Notation. A notation in which the quad- 
rant is divided into 100 parts, called grades, the grade into 
100 minutes, and the minute into 100 seconds, has been in- 
troduced by the French. Its use has met with but little en- 
couragement, owing to the fact that almost all mathematical 
tables are computed upon the other basis. 

9. Circular Notation. This notation is used exclusively 
in the higher branches of mathematics for the measurement 
of angular quantity. 

If we measure the arc of a semi-circle, using an arc of the 

same circle equal in length to the radius as our unit, we find 

-that the arc contains the unit three times and a little more. 

Or, to be exact, since we have from geometry : 

circumference = 2-r, 

and semi-circumference = ttt, 

we see that the semi-circumference contains the unit n (== 

3.14159....) times. This relation is obviously independent of 

the size of the circle. The angle subtended by an arc equal 

to the radius is therefore always the same, being equal to 

180° 

, and this angle is the unit in the circular notation. It is 

called a radian. Hence we write 180° = tt radians, 90° = \tz 
radians, etc. ; or, as the word radian is generally omitted, we 
say simply : 



*■} (1) 



12 PLANE TRIGONOMETRY. 

360° == 2tt, 180° = ~, 90° = 

45° = i-, 30° = l-Tz, etc., 

where we must understand the equality sign to indicate equiv- 
alence, and not identity. 

If we perform the division indicated above (180° -f- -k) we 
find for the equivalent value of 1 radian, the following : 

1 radian = 57°.296 \ 
or 1 radian = 3437'.75 I (2) 

or 1 radian = 206264".8 J 

To determine the number of radians in an arc whose length 
and radius are given, we have from what goes before the 
formula 

= 1, (3) 

where 6 represents the circular measure of (number of radians 
in) the angle which the arc subtends, I the length of the arc, 
and r the radius. Or if 6 and r are given, to find I, we have 

1 = rd. (4) 

10. To pass from one notation to the other. We have just 
seen in § 9 that 

1 radian = degrees. 

And hence 1° = — — radians. 

180 

"We have therefore the following rules : 

180 
I. To reduce radians to degrees multiply by . 

180 
II. To reduce degrees to radians divide by . 

*log 180 = 2.25527 
logTr = .49715 



log — = 1.75812 



* The student is supposed to be familiar with the use of the tables of 
logarithms of numbers. 



THE MEASUREMENT OF ANGLES. 13 

This logarithm will be found convenient in making these 
computations. 

EXAMPLES. 

1. Upon a circle, radius 13 feet, find the length of an arc of 
126° 12'. 

We first reduce the degrees to radians, by Rule II., and then 

find the result by applying (4). 

12G° 12' = 12G°.2 log = 2.10106 

180 
Call the angle in radians, 0. log — = 1.75812 



By Rule II., log = .34294 

lo^r =1.11394 



. • . By (4) log I = log rd = 1 .45688 

.'.1 = 28.634 feet. 

2. Upon a circle, radius 100 feet, how many degrees and min- 
utes in an arc whose length is 1000 feet? 

We first reduce the given arc to radians by (3). The operation 
in this case does not require the use of logarithms, since we have 
at once 

0= I = iopo =lo . 

r 100 
We then apply Rule I. The use of logarithms can be dispensed 
with here also ; and we have 

6 = 10 radians = 10 X — degrees. 

7T 

= 10X57°. 296. 

= 572°.96 = 572° 57 / .6. 
That is, the arc in question is a whole circumference and 212° 
57' over. Now, it is obvious that in this connection these two 
angles do not mean the same thing, and that without using an 
angle greater than 360° we could not solve this problem at all. 
Hence the student will see the importance of the general concep- 



EXAMPLES. 

1. Construct the following angles and tell the quadrant of each 
one : 120°, 215°, 75°, —165°, 300°, 480°, —620°, 620°, —750°, —840°, 
780°, 1000°, 1260°, —1470°, 1530°. 

2. Express the following angles in radians, giving the results 
as multiples of it : 15°, 45°, 60°, 90°, 135°, 210°, 270°, 285°, 300°, 375°, 

2 



14 PLANE TRIGONOMETRY. 

450°, 480°, 585°, 600°, 675°, 720°, —225°, —75°, —105°, —120°, —180°, 
—330°, —495°, —630°, -690°, —750°. 

3. Reduce the following angles to degrees : §7r, |tt, ^r, J^r, |tt, 

§7T, fTT, |7T, £7T, — ^7T, — §7T, StT, — ^TT, (|tT + f 7V — >") , (JtT — |tT + |tt), 

(1^-tV)-(^-^)- 

4. Tell the complement and supplement of each of the following 
angles : 12°, 38°, —75°, 121°, —137°, 156°, 212°, —396°, 428°, £*•, |tt, 

f^, — f *•. I 71 ") ¥*"» — I 71 "- 

5. Find the circular measure of the following angles : 

(1) 13° 19'.6. (3) 173° 8'.3. (5) 378° 16'. 

(2) 67° 18'.1. (4) 296° 13'. (6) 576° 52'. 

6. Find the length of the arc subtending (1) above, if r = 6 feet ; 
of that subtending (2), if r = 7.83 feet. 

7. Find the number of degrees and minutes in the following 
arcs (r = unity) : .1763, .6241, 1.6389, 2.7461, 3.8796, 5.2411. 

8. The lengths of two arcs are 1792.8 feet and 563.87 feet, and 
they form parts of circles of radii 462 feet and 500 feet respectively. 
How many degrees and minutes in each ? 

9. Find the circular measure of an angle of 1°, of 1', of 1". 

10. Two men run in the same direction in a circular track of 
radius 210 feet, one at the rate of 15 feet per second and the other 
at the rate of 16J feet per second. At the end of 25 minutes how 
many degrees and minutes will the second be ahead of the first? 

11. Show that in circles of different radii the number of degrees 
in arcs of equal lengths are inversely proportional to the radii. 



CHAPTEE II. 



THE TRIGONOMETRIC FUNCTIONS. 



Fig. 4. 



11. The Trigonometric Functions. Let A, Fig. 4, be 
any angle less than 90°. From 
any points, B, B v B 2 , B v etc., in 
one of the sides of the angle let 
fall perpendiculars, BC, B X C V B 2 C 21 
B 3 C V etc., to the other side. The 
triangles ABC, AB X C V AB 2 C 2 , etc., 
are all similar ; hence the ratios 
between their sides are equal, and 
Ave have : 

BC 




AB 

AC 



AB, 






= M = ^ = etc, 
AB AB X AB 2 



BC 

AC 



BA 



= M2 = etc. 



AC AC, 



The student will observe that altogether it is possible to form 
six such sets of equal ratios, which consist of the three given 
above and their reciprocals. Thus : 

BC AC BC AC AB^ AB . 

AB' AB' ~AC' BC' AC' BC ' 

and these six ratios have always the same value wherever the 
perpendicular is drawn. They depend, in fact, only upon the 
size of the angle A. For this reason they are called the trigo- 
nometric functions of A. When their values are known the 
angle is known, precisely as if the number of degrees which 
it contains were known. These ratios or trigonometric func- 

15 



16 



PLAXE TRIGONOMETRY. 



Fig. 4. 



tions are known respectively as the sine, cosine, tangent, cotan 
gent, secant, and cosecant of the 
angle A. 

12. Definitions of the Trig- 
onometric Functions. The 
sine {sin) of an angle equals the 
perpendicular let fall from any 
point of one side upon the other 
side, divided by the distance from 
the vertex to the point from 
which the perpendicular was dropped. 

BG 

AB 




C O x C 2 G 3 



Thus, Fig. 4 



sin A 



Mi 

AB, 



etc. 



(5) 



The cosine (cos) of an angle equals the distance from the 
vertex to the foot of the perpendicular described above, 
divided by the distance from the vertex to the point from 
which the perpendicular was dropped. 



Thus, 



cos A 



AC AG 



= etc. 



(6) 



AB AB X 

The tangent (tan) of an angle equals the perpendicular, 
divided by the distance from the vertex to the foot of the 
perpendicular. 

Thus, tan A = ^ == ^ = etc. (7) 



AG AG 



The cotangent (cot) of an angle is the reciprocal of the 
tangent of the angle. 



Thus, 



cot A 



AG __AG X 
BG ~ B,G 



etc. 



(8) 



The secant (sec) of an angle is the reciprocal of the cosine 
of the angle. 

Thus, sec A = ^- = A5l = e tc. (9) 

AG AB X 

The cosecant (esc or cosec) of an angle is the reciprocal of 

the sine of the angle. 



Thus, 



esc A = 



AB AB, 



BG 



*A 



etc. 



(10) 



THE TRIGONOMETRIC FUNCTIONS. 17 

Besides these six functions there are two others of less fre- 
quent occurrence, called the versedsine (vers) and coversedsine 
(covers). They are defined by the following equations : 

vers A = 1 — cos A. (11) 

covers A = 1 — sin A. (12) 

13. The definitions of the cot, sec, and esc above, give rise 
to the following equations, which the student should memo- 
rize : 

cot .-1 = -, tan A = — — , tan A cot A 

tan A 

sec A = -, cos A 

cos A 

esc A = , sin A = 



sin A esc A 

14. Draw a right triangle, 
ABC, Fig. 5, in which a repre- 
sents the side opposite A, b the 
side opposite B, and c the hy- 
pothenuse. Then, according to 
the definitions, § 12, we have 




1 fl 

sm A = — , 
c 






T> CI 

cos B = — , 
c 


cos A = — , 
c 






sm B = — ; 
c 


sin A = cos B. 


ai 


id 


cos A = sin 



but B = 90° — A, 

. • . sin A = cos (90° — A), and cos .-1 = sin (90° — A). (16) 

In the same manner the student can show that 
tan^=cot (90°— A), cot A = tan (90°— A). (17) 

sec A = esc (90° — .1), esc A = sec (90°— A). (18) 

vers A = covers (90° — A). covers J.=vers(90°— A). (19) 
That is : Any trigonometric function of an angle is equal to the 
co-function of the complement of the angle. 

15. The student should be very sure that he understands 
the nature of these quantities. The trigonometric functions 

b 2* 



18 PLANE TRIGONOMETBY. 

of an angle, being ratios of two lines or linear distances, are 
themselves pure numbers. Thus, if in a right triangle ABC we 
have a = 3 inches, 6 = 4 inches, c = 5 inches, then 

sin A = | , tan A = f , sec A = -f , etc., 
sin B = -|, tan 5 = |, sec B = -J, etc. 

Again, in a right triangle in which a — 5 inches, b = 12 inches, 
and c = 13 inches, we have 

sin A = T \= .3846 = cos B, sin B = {% = .9231 = cos A, 
tani = ^= .4167 = cot B, tan B = ±£ = 2.4000 = cot A, 
sec A = if = 1.0333 = esc 5, sec B = ±J-= 2.6000 = esc A. 

We might in this way multiply examples indefinitely, and 
they would all tend to show that each time we take a new 
set of sides for our triangle the new angles thus formed have 
a special set of numbers associated with them as their sin. 
cos, tan, cot. etc. When any one of these numbers is known, 
the angle corresponding to it can be found, and vice versa. 
For this purpose tables giving the values of the functions of 
all angles and also the logarithms of the functions form an 
important part of every set of mathematical tables. 

16. EXERCISE. 

The student can readily construct an experimental table, giving 
the functions for every 5°, as follows : Carefully lay down a line, 
AC, Fig. 6, of any length, say 10 
inches. At C erect a perpendicular Fig. 6. 

of indefinite length, CD. At A, 
with a protractor, lay down an 
angle of 5°, CAB X . Carefully meas- 
ure ^Cand AB V Then by (4), (5), 

(10), the values of the functions 

of 5° and of 85° {AB X C) can be 

found. Next construct a new angle, 

CAB 2 = 10°, and repeat the same operations upon this triangle, 

thus determining the functions of 10° and 80° (AB 2 C). When 45° 

is reached the table for the quadrant will be complete. By using 

care in constructing the figure and in making the measurements, 

the results ought to be accurate to two or three places of decimals. 




THE TRIGONOMETRIC FUNCTIONS. 19 

17. Elementary Relations between the Functions. 
In every right triangle whose sides and hypothenuse are 
respectively a, b, and c, we have 

a 2 -f b 2 = c 2 . (a) 

If this equation (a) be divided successively by the three quan- 
tities c 2 , b 2 , and a 2 , we obtain the three equations 

|+i=7> w 

i-f- = -. (<n 

Now, from Fig. 5 we have by (5) and (6), 

<7 2 & 2 

— = sin 2 A. and — ■ = cos 2 A : 

c 2 e- 

. • . 03) becomes sin 2 A -f cos 2 A = 1. (20) 

Similarly, by ( r ), (7) and (9), 

tan 2 A + l = sec 2 A, (21) 

and by (<S), (8) and (10), 

cot 2 A + l = esc 2 J.. (22) 

Again, by. (5) and (6), we see that 

sin A a b a c a 

X 



cos Ac c c b b 
a 
b 



but, by (7), ~ = tan A 



(23) 



Hence, tan A = . 

cos A 

Hence, also, by (13), cot A = . 

J K sin .4 

Eemark. — The equations just derived, (20), (23), with 

(13), (14), (15), are the most fundamental formulas of trigo- 
nometry. They tell us all the most important relations that 
exist between the different functions of every angle, and for 
this reason should be carefully memorized by the student. 



20 PLANE TRIGONOMETRY. 

18. To determine the numerical Fig. 7. 

values of the functions of 30° and 
60°. 

Let ABC be a right triangle ^r- 

in which A = 30° and B = 60°. /^ 

Draw AB V making an angle of 30° ^_ 

with AC, and prolong BC until it ^ \ 

meets AB X in the point B v ABB X v \ 

is now an equilateral triangle, and 
BC=iBB 1 : 

BC=\AB, 
or AB = 2BC 

By the well-known theorem of right triangles, 



AC= V AB 2 — BC 2 = V±BC 2 — BC 2 = V'SBC 2 : 

AC=BCy3. (0) 

We then have by (a) and (/3), and 

by (5), sin 30° = g = ^= 1 = cos 60° ; (24) 

by (6), cos 30 ° = ^f=f^ = ^T/ 3 - sm 60°; (25) 

by (23), tan 30° = 8m 30 ° = -i- = — == cot 60° ; (26) 

Jy h cos 30° V 3 l/ 3 

by (13), cot30 o = t -^ = -i- = v /3 = tan60°; (27) 

T 3 

by (14), sec 30° = — = — = — = esc 60° ; (28) 

y K h cos 30° V 3 V s . 

by (15), esc 30° = — - — = — = 2 = sec 60°. (29) 

J K J sin 30° I K } 

19. To determine the numerical values of the functions of 45°. 

We can here make use of the fact that 45° is the comple- 
ment of itself. By (16) we have 

sin 45° = cos 45°, 
and by (20), sin 2 45° -f cos 2 45° == 1. 



THE TRIGONOMETRIC FUNCTIONS. 21 



Combining these two equations, we have 




2 sin 2 45° = 1, 




and also 2 cos 2 45° = 1 ; 




whence sin 2 45° = cos 2 45° = -, 




and sin 45° = cos 45° = Jl = X j/2 ; 


(30) 


by (23), tan 45° = S1D 45 ° = 1 = cot 45° ; 
J K h cos 45° 


(31) 


by (14), sec 45°— 1 - — j/2 — esc 45°. 


(32) 



These same values may be found directly from an isosceles 
right triangle by means of the definitions of the trigonometric 
functions. Let the student derive them thus : 



20. EXERCISES. 

4 
1. Given tan A = -p-, to find the other functions of A. 

5 

By (13), cot^l = — "— 7 = -L = 4-. 

J v " tan A 4 4 

IT 
By (21), sec 2 ^ = l + tan 2 ^ = l + ^ = |i: 

sec A = -=- t/41. 
By (22), csc 2 ^ = l + cot 2 ^ = l + ^ = ^: 

cscA = -^-i/41. 
By (15), gin .4 = -J- =1-. = -^. = £•«., 

By (14), cosJ. = 7 = ^ =-7TT = 7Tl/41. 

J v ; ' sec A 1 .. n |/41 41 v 

yl /41 

What we have here accomplished, starting with tan A as the 
known function, can be done equally well if we start with any 
other function as the known one. 



22 PLANE TRIGONOMETRY. 



2. Given tan A = 


— , to find s] 

?2 


As in Ex. 1, 


cot ^4. 


sec 2 A - 


= 1 + tan 2 A 


esc 2 A 


= 1 + cot 2 ^ 




m 2 


esc 2 A 


m 2 -f- n 2 


... z\nA- m 





m ' 




m 2 


m 2 -f n 2 


1+ ™ 2 " 


7l 2 ' 


1 + n% 


m 2 + « 2 t 


m 2 


m 2 


cos 2 .4 


1 n 2 


sec 2 ^1 m 2 -f n 2 


r»os A 


n 



Vm 2j rn 2 V m 2 + n 2 

These results are of sufficient importance to be worth memo- 
rizing. The mathematician frequently has occasion to express the 
sine or cosine of an angle when the tangent is given, and it is 
inconvenient to go through the process of deriving the values 
each time. 

3. Given cos A = 3 sin A, to find the values of the functions 
of A. 

In order to solve this problem we must derive from the given 
equation another that will contain only a single function of A. 
This can be done most conveniently here by substituting for 
sin A its value V\ — cos 2 A derived from (20). This gives us 

cos A = 3Vl — cos 2 A : 
whence cos 2 A = 9 — 9 cos 2 A, 

and 10 cos 2 A = 9: 

cos .4 = -!-. 
t/10 

Hence we have 



sin A = y 1 — cos 2 ^ = V ! — 9 



10 \ 10 t/10 



gob A 3 

esc A = — - — = i/10, 
sin^l 



1 



tan A 



sin A t/10 



cos A 3_ 3 ' 

t/10 



cot A = — - — = 3. 
tan A 



THE TRIGONOMETRIC FUNCTIONS. 23 

4. In a right triangle ABC, Fig. 5, let tan ^4 = ^ and c = 2G 
inches. Find the lengths of a and b and the tangent of B. 
Knowing that tan A = ' , we must have 







sec A — 


Vh 


25 
144" 


1109 
~Vl44- 


13 . 
12 ' 


but 




sec A = 


c . 
6 




. c 

/> 


13 
12 ' 


from which, since c = 26 


inch 


es, we 


have 






r, 


26 __ 
b 


13 

" 12' 


or b = 


= 24 inches. 


Nov 


a = V '(?- 


-6 2 = 


:l/57fc 


— 476 = 1 


100: 


.•. 






« = 


10 in 


ches. 




Hen 


ce, finally, 


tan B = 


b 
a 


24 _ 12 
10 5 * 




5. 


Prove tin 


, sec A + esc A _ 


1 -foot A 





sec A — esc A 1 — cot A 
First Solution. — We can show by successive changes of the 
expression on the left that the two are equal. 

Substitute for sec A, and for esc A : we then have 

cos A s'mA 

cos A am A 1 -4- c ot A 

1 1 _ 1 — cot A' 



cos A sin A 
Multiply numerator and denominator on the left by cos A ; we 
then have 

1 _i_ cos ^4 

sin A 1 + co t A 

-j cos A 1 — cot A? 

sin A 

which we see is true, since -^ — = cot J. 

sin A 

Second Solution. — We can show by successive changes of the 

expression on the right that the two are equal. 

Substitute ^- s -^ for cot A ; we then have 
sin A 

, , cos A 



sec A + esc A _ sin A 



sec A — esc A ^ cos A 

sin A 



24 PLANE TRIGONOMETRY. 

Divide numerator and denominator on the right by cos A ; we 
then have 

sec A -f- esc A cos A sin A 

-i t i 



sec A — esc A 



cos A sin A 



which we know is true, since sec A = , and esc A = . 

cos A sin A 

Third Solution. — Multiply both terms of the fraction on the left 
by sin A cos A, and both terms of that on the right by sin A ; 
remembering that 

sec A = — - 1 — , esc A = — - 1 — , and cot A = ™®A., 
cos A sin A , sm^L 

we have as our result 

sin A -\- cos A _ sin A -f cos A 
sin A — cos A sin A — cos A' 

This is an identity ; and it shows the truth of the equality from 
which it was derived, because the operations that were performed 
on the two sides of the given equation left both of them un- 
changed in value. 

It has been shown, in what has gone before, how from the 
known lengths or ratios of lines the values of the functions of 
angles can be determined. The following ex- 
ample will show how from a given line and 
given angles the lengths of other lines may be 
determined. 

6. In Fig. 8 let POM=x and MON=y be 
known angles, and OP a known distance. The 
angles OPM, OMN, and MRN are supposed 
to be right angles. It is required to find the 
lengths of all the lines in the figure in terms 
of OP and the two given angles. 
Solution : 

MP=OPtfmx. 
031= OP sec x. 

MN= OM tan y = OP sec x tan y. 
ON= OJ/sec y = OP sec x sec y. 
MP = OM sin y = OP sec x sin y. 
If we suppose the length of OP to be 12 inches and the angles 
x and y to be 30° and 45° respectively, we have 

MP = 12 tan 30° = 12 • -i- = 4 ^3 inches. 
>/3 




THE TRIGONOMETRIC FUNCTIONS. 25 

OM= 12 see 30° = 12 • — = 8 |/3 inches. 
l/3 

MN= 12 see 30° tan 45° = 12 • - 2 - • 1 = 8 |/3 inches. 

l/3 

OJV= 12 sec 30° sec 45° = 12 • -£- • i/2 = 8 /6 inches. 

1/3 

J/# = 12 sec 30° sin 45° = 12 • — • - 1 1/2 = 4 ,/6 inches. 

l/3 2 1 

EXAMPLES. 

Id the following right triangles write out the values of all the 
functions of A and B: 

1. a = 21, 6 = 20. 3. a = 7, 6 = 5. 

2. a = 15, 6 = 20. 4. a =11, 6 = 3. 

In 4 reduce the results to decimals and carry them out to three 
places. 

In the following examples find from the function given in each 
case the values of the other five : 

5. tan A = -. 8. cos A = J — . 11. sin A = .46. 

2 A7 

6. sin^l = -— . 9. cot .4 = -. 12. tan ^4 = m. 

1 3 9 

7. sec ^4 = -. 10. sec ^4 = 1.92. 

5 

From each of the six equations following determine the values 
of the six functions of the angle : 

13. sec .4 = 3 cos A. 15. tan ^4 = } see X 17. tan A = m sin A. 

14. sin A = | cos A. 16. tan ^4 = 5 sin A. 18. sec A = n tan ^4. 

19. Prove that 1 + sin ^4 = 



1 — sin A 

i 
20. Prove that sec A -4- tan ^4 



sec ^4 — tan ^4 

21. Prove that cot ^ cos ^ ^ cot ,4 - cos ^ 
cot A -f- cos A cot ^4 cos A 

22. Prove that tan A + cotA = sec2 ^ + csc ' A . 

sec ^4 esc ^4 



26 



PLANE TRIGONOMETRY. 



Fig. 9. 




Fig. 10. 



23. Prove that (r cos x) 2 -\- {r sin x sin y) 2 -\- (r sin x cos ?/) 2 = 

24. In a right triangle given a = 7 inches and sin i? = f 
6, c, and tan A. 

25. In a right triangle given c = 12 
inches and sec J. = 3 : find a, 6, and 
cos _B. 

26. In a right triangle tan B = f and 
c = 4 inches : find a, b, and sin .4. 

27. In Fig. 9, OP =6 inches and 
PON = 60°. O.l/P, O/W, and OiVi? 
are right angles. Find the lengths of 
ON, PM, PN, NR, and OR. 

28. In Fig. 10, OP = 12 inches, POM 
= 310 N = NOR = 30°. The angles 
OPM, OQP, OMN, OSM, ONR, and 
OTN are all right angles. Find the 
lengths of OM, ON, OR, PM, MN, NR, 
PQ, MS, and NT. 

29. In Fig. 11, given OP = 6 inches, 
POR = NST = 60°, and MRT = 45°. 
PRMN is a square. Find the lengths of 
OR, PR, RT, NT, TS, and OS. 

Find by the use of the logarithmic 
tables the values of the following ex- 
pressions : 

2.46 X sin 78° 12' 61.74 X tan 2 36° 12' 

6.38 X tan 17° 9'' 19.45 X cos 39° 14' 



find 




30. 



31. 



32. 



Z4.96 X tan 63^ X cos 2 9° m i = A pind A 
V63.71 X sin 3 41° 38' X cot 26°/ 

(478.9 X cos 83° 6' X tan 2 79° 38QI 
(cot 3 31° 16' X sin 81° -=- sin 2 41° 36')*' 



21. Trigonometric Functions of any Angle. In 

Chapter I. an angle was defined in such a way as to admit of 
an unlimited numerical measure ; but so far the trigonometric 
functions of an angle have been defined with reference only 
to angles less than 90°. We shall show how these definitions 
apply to all angles. 

Let the angle ACP V Fig. 12, be represented by V ACP 2 by 



THE TRIGONOMETKIC FUNCTIONS. 



27 



Fig. 12. 

B 



r ACP 3 by 5 , and ACP± by V the angles all being measured 
in the positive direction, as in §2. 
This notation is adopted merely 
as a matter of convenience, so 
that the subscript of 6 may indi- 
cate the quadrant to which the 
angle belongs. If we write 
without a subscript it will mean 
that the angle is not limited to 
any particular quadrant. If the 
angle is 6 V it is an acute angle, 
ACP 1 for example, and from the 
definitions (§ 12) we have 



p 




c/ 




*i 


4, 












w 2 / 




M, 




P 3 








P 

4 



B, 



sin 0, 



3LP, 



cos l = 



CM, 



tan 0, 



^A 



CP^ l - CP l ' l CM, 



etc. 



If the angle is greater than 90°, our definitions still apply, but 
in the case of some such angles the side to which the per- 
pendicular is drawn must be extended backward through the 
vertex of the angle in order to make the necessary intersec- 
tion with the perpendicular. This must be done if the angle 
is ACP 2 (0 2 ), or ACP 3 (0 3 ), where the side CA must be pro- 
duced to the left, through C, to intersect the perpendiculars 



let fall from P 2 and P 3 . 



• a M 2 P 2 



cos O 



We then have 
CJL 



CP n 



tan O 



M 2 P 2 
CM 



etc. 



sin 0, = 



CP> 



COS O 



CM n 



tan 



M 2 P 3 



etc. 



CP 3 7 " CM 2 ' 

For an angle in the fourth quadrant ACP± (0 4 ), we have 

CM, . „ 31 P, 



sin 0. 



CP, ' 



cos 6 A 



CP. 



tan 6 t 



CM, 



etc. 



As it will be convenient to have single letters to stand for 
the three lines which enter into the trigonometric ratios of 
any angle, we shall always designate the length of the perpen- 
dicular by y, the distance from the vertex of the angle to the 



28 



PLANE TRIGONOMETRY. 



Fig. 12. 



foot of this perpendicular by x, and the distance from the 

vertex of the angle to the point from which the perpendicular 

is dropped by r. We shall further regard y as positive when 

the angle is between 0° and 180°, and negative when the angle 

is between 180° and 360°. The reason for this is plain. Ee- 

ferring to Fig. 12, it is seen that 

when the angle has one side in 

coincidence with CA, then for 

angles in the first and second 

quadrants y is measured upward, 

and for angles in the third and 

fourth quadrants y is measured 

downward. Thus y is measured 

sometimes in one direction and 

sometimes in the other, and this 

difference in direction must be 

indicated by difference in sign. 

For the same reason x is taken as positive when the side to 

which the perpendicular is drawn is not produced through the 

vertex to meet the perpendicular, and negative when this side 

is so produced. Since r is merely a distance, with which the 

idea of direction is not associated, as in the case of x and y, it 

will be always positive. 





B 






p 2 








P i 


A. 




c/ 






*H *• 


M 2 / 




M, 




P 3 








P < 



Bi 



Using this new notation, we can write 



sin 



r 



(33) 



cos = . 

r 



tan 



cot 



sec = — . 
x 



esc 



(34) 
(35) 
(36) 
(37) 
(38) 



THE TRIGONOMETRIC FUNCTIONS. 



29 



Note. — On the graphical repre- 
sentation of the trigonometric func- 
tions. 

If we denote the angle ACP 
by 0, we may write, according to 
the definitions, 



sin 



PM 



CP 



sec 



tan 



CT 



AT 



CA 
cot = tan BCT 



BE 




CA 



cos = sin BCT = 4~ = 



CB' 

where it will be observed that the radius forms the denomi- 
nator of each fraction. 

Now, if the radius be taken as the unit of length, it follows 
that 



sin 
costf 



PM, 



tan = AT, 



= CM, cot = BR, 



sec = CT, 
esc = CB. 



We do not mean that the functions are lines, but only that 
under some circumstances they may be represented by them. 
It is from this, the original, method of defining the trigono- 
metric functions that their names are derived. 

When, therefore, the student meets with the statement that 
such and such a line is the sine or cosine, etc., of such and such 
an angle, as he may do in some of the older writings where 
trigonometry is applied, he will be prepared from what is said 
above to understand what is meant. 

22. Elementary Relations. The formula established 
in §| 13 and 17 were derived for angles less than 90° ; but it 
is easily seen that they are equally true for the functions of 
any angles. (13), (14), and (15) must be true for all angles, 
for they follow directly from the definitions of the functions. 
To extend the application of (20), (21), (22), and (23) to 

3* 



30 PLANE TRIGONOMETRY. 

angles greater than 90°, we proceed as follows. Whatever 
may be the size of the angle 6, Fig. 12, we must have 

x 2 -\- y 2 = r 2 : 

, x 2 , y 2 n 

hence _ + ^. = i ; 



. • . by (33) and (34), sin 2 6 -f cos 2 = 1. (20)* 

The student can then readily show that 

tan 2 + l = sec 2 6>, (21) 

cot 2 + 1 = esc 2 6, (22) 

, Q sin , . cos0 /OON 

tan = , cot = . (23) 

cos# sin d 

23. Signs of the Functions in Different Quadrants. 

In § 21 it has been explained when x and y are positive and 
when negative; we shall now show what will be the effect of 
this convention of signs upon the trigonometric functions. 

Referring to Fig. 12, the student will see that for all angles 
in the first quadrant both x and y are positive. It follows, 
therefore (since r is always positive), that all the functions of 
angles in the first quadrant are positive. 

In the second quadrant x is negative and y is positive. 
Hence those functions which contain x will be negative, and 

the rest will be positive. From (33) (38) we see, therefore, 

that cos, tan, cot, and sec will be negative, and sin and esc 
positive. 

In the third quadrant both x and y are negative. In this 
quadrant, then, we shall have sin, cos, sec, and esc negative, 
and tan and cot positive. 

In the fourth quadrant x is positive and y is negative. 
Hence in this quadrant the negative functions will be sin, tan, 
cot, and esc, and the positive ones will be cos and sec. 

These results are summarized in the following table: 

* This and the three following formulas are numbered to correspond 
with those' in \ 17, with which they are identical. 



THE TRIGONOMETRIC FUNCTIONS. 



31 



Angle = 


0°-'J0°. 


90°-180°. 


180°-270°. 


270°-360°. 


sin . . 


+ 


+ 


— 


— 


cos . . 


+ 


— 


— 


+ 


tan . . 


+ 


— 


+ 


— 


cot . . 


+ 


— 


+ 


— 


sec . . 


+ 


— 


— 


+ 


CSC . . 


+ 


+ 


— 


— 



24. Variations in the Value of the Sine as the 
Angle increases. 

In Fig. 1-4 let us imagine a line, Fig. 14. 

CP, to revolve about C in the 
positive direction as explained in 
§ 2. Let the angle it makes with 
GA be called 0. Before CP com- 
mences to revolve it will coincide 
with CA, and in this position is 
said to make an angle of 0° with 
CA. The value of y will then be 
0, and we shall have 



sin 0° = 



y 



o 



= 0. 




r r 
As the radius revolves from CA to CB through the first 



quadrant y will continually increase, and hence Jl or sin 



will increase. When CP reaches the position CB we have 
y = r and = 90°. 

ii r 
sin 90° = — == — =1. 



Following the revolution of CP on through the second 
quadrant, we see that here y continually decreases, and hence 

}L or sin 6 will decrease. When CP reaches the position CA V 
r 

y = Q and 0=180°. 

if 

sin 180° = — = - 



0. 



32 PLANE TRIGONOMETRY. 

As the radius CP revolves through the third quadrant y 
increases numerically, but, being negative, decreases in an 
algebraic sense. Hence in the third quadrant sin 6 decreases. 
When CP coincides with CB V y = —r and = 270°. 

V — r 

sin 270° = — = = — 1. 

r r 

As the radius CP revolves through the fourth quadrant 
y decreases numerically, but, being negative, it actually in- 
creases, so that in the fourth quadrant sin d increases. When 
CP reaches CA we have again y = and 6 = 360°. 

y 
sin 360° = — = — = 0. 
r r 

It is evident from this investigation that the value of the 
sine of every angle lies between -j-1 and — 1. 

Remark. — We might follow the revolution further and con- 
sider angles greater than 360°, but it is easy to see that this 
would lead to no additional result be} T ond the fact that the 
sine of any angle "Inn -j- 6 (0 being <360°) is always the same 
as sin 0. This remark applies as well to the investigations 
that follow as to the investigation of the sine given above. 

25. Variations in the Value of the Cosine as the 
Angle increases. Without giving such a detailed descrip- 
tion of the process as that above, the student will be able 
readily to deduce the following facts in regard to the cosine. 
When = 0°,x = r. 

cos0° = — = — = 1. 

r r 

In the first quadrant, while d increases, x decreases, and 

hence — or cos 6 decreases. When = 90°, x = 0. 
r 

cos 90° = — = — = 0. 
r r 

In the second quadrant x decreases algebraically as 6 in- 
creases. Hence cos 6 decreases. When d = 180°, x = — r. 

cos 180° = — = — =—1. 
r r 



THE TKIGONOMETRIC FUNCTIONS. 33 

la tho third quadrant x increases algebraically as in- 
creases. Hence cos increases. When = 270°, x = 0. 

cos270°= — = — = 0. 
r r 

In the fourth quadrant x continues to increase as in- 
creases. Hence cos 6 increases. When = 360°, x ^= r. 

cos 3G0° = £ = ~ = l. 
r r 

We see from this that the values of the cosine always range 
between -j-1 and — 1, the same as those of the sine. 

26. Variations in the Value of the Tangent as the 
Angle increases. Taking up the investigation of the tan- 
gent, we see from (35) that 

tan 0° = — = 0. 
r 

In the first quadrant y increases and x decreases. Hence 
tan d increases. At 90°, y = r and x = 0. 

tan 90° = — = oo. 


In the second quadrant y decreases and x increases numeri- 
cally, but is negative. Hence tan decreases numerically, 
but, being negative, increases algebraically. When 0= 180°, 
y = and x = — r. 

tan 180° = — = 0. 
— r 

In the third quadrant y increases numerically and x de- 
creases numerically, but both are negative. Hence tan 
increases. When = 270°, y = — r and x = 0. 

tan 270° = — = oo. 


In the fourth quadrant y decreases numerically and is nega- 
tive, while x increases and is positive. Hence tan increases 
algebraically. When = 360°, y = and x = r. 

tan 360° = — = 0. 
r 



34 



PLANE TBIGONOMETRY. 



We thus see that the range of values of the tangent is 
from — oo to -j-oo. Hence there is no number that is not the 
tangent of some angle. This is frequently a very convenient 
fact. 

27. The student should repeat the same course of reasoning 
for the remaining three functions, — cotangent, secant, and 
cosecant. The facts concerning them can easily be deduced 
from their reciprocal relations with the tangent, cosine, and 
sine respectively. 

The above results are summed up in the following table : 





0°. 


First 
Quad- 
rant. 


90°. 


Second 
Quad- 
rant. 


180°. 


Third 
Quad- 
rant. 


270°. 


Fourth 
Quad- 
rant. 


360°. 


sin . . 





+, Inc.* 


1 


+, Dec. 





— , Dec. 


— 1 


— , Inc. 





cos . . 


1 


4- , Decf 





— , Dec. 


— 1 


— , Inc. 





+, Inc. 


1 


tan . . 





+, Inc. 


00 


— , Inc. 





-f , Inc. 


00 


— , Inc. 





cot . . 


00 


+, Dec. 





— , Dec. 


00 


+, Dec. 





— , Dec. 


00 


sec . . 


1 


+, Inc. 


00 


— , Inc. 


— 1 


— , Dec. 


00 


-f , Dec. 


1 


CSC . . 


CO 


+, Dec. 


1 


+, Inc. 


00 


— , Inc. 


— 1 


— , Dec. 


00 



EXAMPLES. 

1. Write the signs of the functions of each of the following 
angles: 178°, 312°, —95°, (f7r + 72 ), (|tt — 13°), 7(^ — 16), (3tt — 
521°), (312° — fO, f(7r + 87°), 529°, 632°, —417°, 896°, 1321°, — 7* -f 
784°. 

2. tan 6 = — f and cos 6 is negative. Tell the quadrant of 6 and 
give the values of all its functions. 

3. sin 6 = — h and tan 6 is positive. Solve in the same manner 
as Ex. 2. 

4. For what values of a is sin a -\- cos a positive, and for what 
values is it negative ? Construct a circle and divide it into sectors 
in which this expression is alternately positive and negative. 

5. Treat the expression sin a — cos a in the same manner. 



6. Treat the expression 



sin a 



in the same manner. 



cos« 



* Increases as the angle increases, 
f Decreases as the angle increases. 



THE TRIGONOMETRIC FUNCTIONS. 35 

7. esc 2 = 10. Show that this equation gives four values of 
less than 360°, one in each quadrant. Find the tangent of each, 
and construct them geometrically. 

8. sin 2 — 3 cos 2 6. Show that this equation gives four values of 
less than 360°, one in each quadrant. Find the sine of each. 
How many degrees in each ? 

9. sin 6 = 3 cos 6. Show that this equation gives two values of 
v less than 360°. Determine their tangents, and construct them 
geometrically. 

10. What series of values of 6 will satisfy the equation 

2tan 2 + l = sec'-tf? 

11. Also sin 2 6 — cos = 1? 

12. Also sin d cos 6 = tan 6 ? 

13. Show that the equation sin 2 -f- cos = 2 is impossible. 

14. Tn each of the following pairs of expressions tell which is 
the greater : 

sin 50° or sin 40°, cos 50° or cos 40°, 

tan 97° or tan 121°, cos 126° or cos 26°, 

sin 37° or sin (—37°), tan 213° or tan 245°, 

tan T 4 2 7r or tan |7r, sin §tt or sin ^-, 

cos 524° or cos 132°, cos (—412°) or cos 241°, 

cot (—129°) or cot (—132°), sin (— S9G°) or —sin 212°. 

15. What are the numerical values of the following expressions ? 

3 sin 90° — 2 cos 180° 
' sec 3(50° ' 

2 sin 270° (tan 180° — 4 esc 90°) 
cos 270° — sec 180° ' 

5 (sin 270° — tan 360° + cos 540°) 
2 esc 270° . sec 900° 

2 sec 540° — 3 sin 810° + 2 cos 630° 

3 esc 1350° -f 7 cos 450° — sec 1260°' 

16. Reduce the following to expressions containing only the 
sine of the an^le : 



U) 


sin a cos 2 a -| 

COS a 




cos a sin a 


(2) 


CSC a sill a -f COS ° - 
COt a 




sin a g 
cos 2 a 


(3) 


tan 2 6 + 1 — esc 2 e -f 


2 


sin 6 cos 6 ; 


(4) 


sec 2 ^ — tan 2 <p -4- cos 2 


<t> 


— cot 2 (p. 



36 



PLANE TRIGONOMETRY. 



17. Reduce the following to expressions containing only the 
cosine of the angle : 

1 , tan a . 



(1) 



sin 2 d . tan 



(2) sec a 



cos cot 

(3) cot 2 x + tan 2 a; 

(4) sin y cos y tan y cot y. 

18. Reduce the following to expressions containing only the 
tangent of the angle : 



(1) 



tan x -f- 1 . 



cot x + 1 ' 

(3) (1 + cos 0) (1 — cos 0) 

(4) sin (f> cos ^ + cos tan 



(2) cot /? + sec |8 - 

f tan (cot 2 — 1) 
A + cot sin 0. 



esc (3 ; 



28. Functions of 90° -f 0, 180° + 0, 270° + 0, in terms 
of d. All mathematical tables give the trigonometric func- 
tions of angles only up to 90°, but in practice we are con- 
stantly meeting with angles greater than 90°. It therefore 
becomes necessary to investigate the relations between func- 
tions of angles greater than 90° and those of angles less than 
90°, so that, knowing the latter, 
we may be able to find the 
former. 

Let ACP V Fig. 15, be any 
angle less than 90°, and let it 
be represented by 6. Draw CP r 
CP V and CP V making with GP 1 
angles of 90°, 180°, and 270° 
respectively. We shall then have 

ACP 2 = 90° + 0, (a) 
ACP 3 =18O° + 0, (/5) 



ACP, = 270° -f 0. 



GO 




It is readily seen that the triangles CP X M V CP 2 M 2 , GP Z M V 
CP±M± are all equal. We have, therefore, 



sin AGP n 



CP n 



CM, 
OR 



cos ACP, 



and cos A CP 2 = — — * 
0P„ 



-MA 

CP 



-sin ACP V 



THE TRIGONOMETRIC FUNCTIONS. 



37 



Hence, by (a), sin (90° + 0) = cos 0, 
cos (90° -j- 0) == —sin 0, 

from which we derive, by (23), (14), and (15), 
tan (90° -f 0) = —cot 0, 
cot (90° + 0) = — tan 0, 
sec (90° + 0) = —esc 0, 
esc (90° + 0) = sec 0. 

In the same way we have in the third quadrant 



(39) 



3 CP S OP, 


-sin AC1 


inn /^r»Q ,4 (!P 3 1 ^^ 


cos ACPj 


3 CP 3 GP X 


Hence, by (/5), sin (1S0° -f 0) = —sin 0/ 




cos (180° + 0) = —cos 0, 




and by (23), (14), and (15), 




tan (180° -f 0) = tan 0, 


■ 


cot (180° -f 0) = cot 0, 




sec (180° -f 0) = —sec 0, 




esc (180° + 0) = —esc 0., 




The proof of the following is left to the student: 


sin (270° -f 0) = —cos 0; 




cos (270° -f 0) = sin 0. 




tan (270° + 0) = —cot 0, 




cot (270° + 0) = —tan 0, 




sec (270° + 0) = esc 0, 




esc (270° -f 0) == —sec 0., 





(40) 



(41) 



29. Function of Angles > 360°. We might continue 
this process and find the functions of (360° -f- 0), but the 
student can see at once that these will be the same as the 
functions of 0, and, in fact, any function of (71 . 360° -f- 0) will 
always be the same as the function of 0. 

Thus, to find tan 2079°, we subtract 5 X 360° = 1800° from 
2079°, thus getting 279°. 



38 



PLANE TRIGONOMETRY. 



Hence tan 2079° = tan 279°. 

Now, by (41), tan 279° = —cot 9°, 

tan 2079°' = — cot 9°. 

30. Functions of 180° — o, 270° — e, 360° — 0, in terms 

of 0. While the foregoing formulae are sufficient to furnish 
any function of any angle in terms of a function of an angle 
less than 90°, it is sometimes convenient to be able to do the 
same thing in another way. For this reason the following 
formulae are given : 

Let ACP 1 be any angle of the 
first quadrant, and let it be rep- 
resented by 0. Construct the 
angle A X CP 2 equal to 0, produce 
P 2 C to P v and draw CP 3 perpen- 



dicular to 



PP. 



The angles 



P 3 CB 1 and P£A will both be 
equal to 0. Hence we have 

ACP 2 = l8O° — 0. (a) 

AGP 3 = 270° — 0. (0) 

ACP, = 360° — 0. {f) 
Equality must exist between the triangles M l CP v M 2 CP 2 , 
M 9 CP„ and M,CP A . Hence we have 




M V MP 
sin ACP 2 - M G *p - M £ - sin AOP v 


and coo iCP CM 2 * -CM,' 


osACP^. 


and cob AGP, _ — _ ^ - o 


. • . by (a), sin (180° — 0) == sin d' 




cos (180° — 0) = —cos 0, 




and by (23), (14), and (15), 




tan (180° — 0) = —tan 0, 


■ 


cot (180° — 0) = —cot 0, 




sec (180° — 0)=— sec 0, 




esc (180° — 0)= esc 0. y 




In the same manner 


. .„- M,P, — CM, 
Bin AGP. = £- cp ;— < 


2osACP v 



(42) 



THE TRIGONOMETRIC FUNCTIONS. 



39 



and cos A (7P, = ■ - - 

3 CP. 



M Jl± = — sin ACP, 



CP, 



. by (/J), and by (23), (14), and (15), 

sin (270° — 0) = — cos 0, 

cos (270° — 0) = —sin 0, 
tan (270° — 0) = cot 0, 
cot (270° — 0) = tan 0, 
sec (270° — 0) = —esc 0, 
esc (270° — 0) == —sec 0. J 

The student can give the proof of the following: 

sin (360° — 0) = — sin 0," 

cos (360° — 0) = cos 0, 
tan (360° — 0) = —tan 0, 
cot (360° — 0) = —cot 0, 
sec (360° — 0) = ' sec 0, 
esc (360° — 0) = —esc 0. 



(43) 



(«) 



31. Functions of Negative Angles. If the angle 
ACP V Fig. 16, be conceived as generated by CP i revolving 
from the position CA in the negative direction, it will be a 
negative angle, as explained in § 5. Let us designate it by 
— 0. Then, if we construct ACP 1 equal numerically to ACP^ 
we have 



sin (—0) 



cos (— 0) 



Mf< 



■M^ 



GP A 



CP, 



-sin 0, 



CM CM, 



i = cos 0, etc. 



CP, CP X 

And we thus derive the formulae 

sin ( — 0) = — sin 0, 
cos ( — 0) = cos 0, 
tan (—0) = — tan 0, 
cot ( — 0) = — cot 0, 
sec ( — 0) = sec 0, 
esc ( — 0) = — esc 0.^ 

32. The results of §§ 28 .... 31 are summed up in the 



(45) 



40 



PLANE TRIGONOMETRY. 



following rules, which the student will find very convenient 
in practice : 



'•J (46) 



I. For angles less than 360° : 
Function (—6), (180° ± 0), (360° — 0) = function 
Function (90° -f 0), (270° ± 0) = co-function 0. 

II. For angles in general : 

Function (jnt ± 0) = function 0. } , ,-. 

Function ( [2n -f 1> ± 6) = co-function 0. ] 

In every case the sign belonging to the function of the original 
angle must be prefixed. 

Thus, if we desire tan 136° 12' 37", we may subtract 90° and 
look in the table for cot 46° 12' 37", or we may subtract the 
angle from 180° and look for tan 43° 47' 23", in both cases 
prefixing the minus sign. The student will see that the first 
method is preferable, because in most cases the subtraction 
can be more easily performed. 

33. The following table contains the functions of a number 
of angles of frequent occurrence, and will be found convenient 
for reference. It is left to the student to show how they are 
all derived from the results of §§ 18, 19, by means of (39), 
(40), and (41), or of (42), (43), and (44). 





sin. 


COS. 


tan. 


cot. 


sec. 


CSC. 


30° 


* 


*l/8 


*l/3 


l/3 


fl/3 


2 


45° 


w* 


W* 


1 


1 


l/2 


l/2 


60° 


h/s 


I 


l/3 


*l/3 


2 


fl/3 


120° 


h/3 


i 

2" 


-l/3 


-*l/8 


—2 


fl/8 


135° 


*l/2 


—w* 


—1 


—1 


-1/2 


1/2 


150° 


\ 


Hh/8 


-h/8 


-l/3 


-It/3 


2 


210° 


i 


— il/3 


V3 


l/3 


-fl/3 


—2 


225° 


-ii/2 


-ii/2 


1 


1 


-l/2 


-l/2 


240° 


— Ji/3 


1 

2 


T/3 


h/3 


—2 


-fl/3 


300° 


— V3 


2 


-l/3 


-Jl/S 


2 


—fl/3 


315° 


— ]l/2 


Jl/2 


— 1 


—1 


l/2 


-l/2 


330° 


2 


21/3 


-4i/3 


-V'3 


.V3 


—2 



THE TRIGONOMETRIC FUNCTIONS. 41 

34. Inverse Trigonometric Functions. It is some- 
times convenient to express the value of an angular magnitude 
in terms of one of its trigonometric functions, instead of in 
degrees or radians. Thus, we may wish to express that a 
certain angle is one whose tangent is f . This determines the 
angle for us just as truly as if we knew the number of degrees 
it contained, except that it leaves us in doubt which of two 
angles, both less than 300°, is meant. The student has found, 
however, that this ambiguity will always arise when all that 
we know about an angle is the value of one of its trigonomet- 
ric functions. To express an angle in this way a special nota- 
tion has been devised. In this notation the angle mentioned 
above would be written 

tan- 1 !. 

This may be read in three ways: "the angle (or arc) whose 
tangent is I," " inverse tangent I," or " anti-tangent f ." These 
three statements, however, all convey precisely the same idea. 
If we choose to designate the angle which we are talking 
about by we write 

tfrrrtan- 1 !, 

which is simply another way of saying 

tan 6 = f . 
The student's attention is called particularly to the two 
equations which follow. These sometimes puzzle a beginner 
when he meets them for the first time ; but a little thought 
will convince him of their truth : 

sin (sin-H) = |-, 

sin - 1 (sin 6) = 6. 

To be complete for angles less than 360° the latter should read 

sin - 1 (sin d) = or 180° — ; 
or, to be perfectly general, we must write 

sin - 1 (sin 6) = 2nit + or (2w -f 1) iz — 0, 

since the expression on the left is satisfied by any angle whose 
sine is equal to sin/?, and of such there is an indefinite number 
if we admit angles of unlimited magnitude. The following 

4* 



42 PLANE TRIGONOMETRY. 

examples will help the student to a clear understanding of 
this notation. It is true that in trigonometry itself it finds 
only occasional application ; but in calculus and the higher 
branches of mathematics it will be met with constantly. If 
the student will remember, when he sees an expression like 
one of those above (e.g., cot" 1 2), that it represents an angle, 
he will have little difficulty with this part of the subject. 

1. What is tan -^y^? 

Evidently the answer is 30° and 210°, because tan 30° = £ -j/3, 
and tan 210° = iV3. 

2. What is tan (cos- 1 !) ? 

Here we are asked to find the tangent of the angle whose cosine 
is f. Let this angle be 8. We then have 

cos = f , 



hence sin = ±l/l — f = =fcj/f = ±J T /5 : 

tan (cos - 1 1) = tan = ±| -/5. by (23) 

6 may be in the first or fourth quadrant. If it is in the first, 
tan 6 = \ j/5 ; 
if in the fourth, tan 6 = — J -j/5. 

3. Show that tan ~ l m + cot ~ l m = ~ or -£. 

A A 

Here we are to show that the sum of two angles is 90° or 270°, 
having given that tan of one and cot of the other are both equal 
to the same quantity m. Let the angles be 6 and 0. Then we have 

tan 6 = m, 
cot (j> — m. 

Hence, by (17) and (43), 6 = 90° — $, or 270° — % 

or 6 + <p = 90° or 270° : 

tan ~ l m 4- cot _1 m = — or ~. 

T 2 2 

EXAMPLES. 

1. Find for each of the following expressions a value in terms 
of the same function of a positive angle less than 90° : sin 197°, 
cos 121°, tan 289°, sec 312°, tan 96°, sin (—128°), cos (—48°), tan |tt, 
cot (— |tt), esc f7r, cot 739°, sin 826°, cos 523°, tan (—702°), sin (—512°), 
sin 426°. 



THE TRIGONOMETRIC FUNCTIONS. 43 

2. Find a value for each of the above expressions in terms of a 
positive angle less than 45°. 

3. Simplify the following expressions : 

(1) a cos (90° — 0) + 6 cos (90° + 0). 

(2) (a — b) tan (90° — </>) + (a + b) cot (90° + 0). 

(3) sin (90° + a) cos (180° — j8) + cos (90° + a) sin (180° — /?). 

(4) cos (180° + a) cos (270° — /3) — sin (180° + a) sin (270° — P) m 

(5) tan a + tan (- (3) — tan (180° + j3). 
,py. n sin « tan (18 0° + a ) 

1 j tan a cos (90° — a) 

m tan (90° + A) cos (270° — 0) cos (— 6) 

{ ' ' " cot (180° + 0) sin (270° + 0) 

4. From the equation sin 2 + cos 2 = 1 show that both sin and 
cos ^ must always lie between +1 and — 1. 

5. From the equation tan 2 + 1 = sec 2 show that sec can 
never be less than unity. 

6. What positive angles less than 8G0° have the same 

tan as 212°, sec as fn-, sin as ^tt, 

cos as —192°, cot as —78°, tan as —119°, 

sin as 47°, cos as 305°, cos as |tt? 

7. Prove the truth of the following equalities : 

(1) cos (90° + a;) cos (1S0°— x) + sin (90° + x) sin (180° + x) == 0. 

(2) sin(450° + a) — sin (270° — a)=sin(450° — a) — sin (270 + a). 
(S) tan (180° — 0) = cot (90° + 0) 

1 ' tan (270° — 0) cot (180° + 0)* 

^ sin (-—</>) + cos (—0) _ sin (90° + f) + cos (270° — <f>) 
{ ' tan (—0) — cot (— <p) cot (180° + 0) + tan (360° — <£)' 
In the examples which follow, as far as 32, find all the possible 
values of less than 360°. 

8. Sin 20 = h 
Solution : If sin 20 = J, 

we have, by (24), (42), and | 29, 

20 = 30°, 150°, 390°, 510°, etc.: 
0=15°, 75°, 195°, 255°, etc. 



9. cos 2 36 = 1. 


11. sin $0 = ^3. 13. tan£0 = V3. 


lO. tan 4^ = 1. 


12. cos-J0=*. 14. sin = cos 78° 


Solution of 14 : 


If sin = cos 78°, 


then, by (16) 


sin = sin 12°. 


Hence 


= 12°, and by (42) = 168°. 



44 PLANE TRIGONOMETRY. 

15. tan ~ —tan 46°. 22. cot = tan (180° + 0). 

1 6. cos = —sin 216°. 23. tan 20 = —tan 14°. 

17. cot0 = tan|7r. 24. cos J0 = —cos 126°. 

18. sin = cos (p. 25. sin 30 = cos 27°. 

19. sin = — cos (180° — <j>). 26. cot $0 = tan 278°. 

20. tan0 = — cot (9O° + 0). 27. cos |0 = sin (— |tt). 

21. cos = sin (270° — 0). 

28 tan — 8 ' 26 X tan 198 ° 1S ' X cos 13 ° 17 ' 

4.76 X sin 28° 16' 

29 sin e — 17 X sin 283 ° 19/ X tan 47 ° 16/ 

39.2 X cos 183° 6' 

30 cot — ~~ 4 - 289 X cos 73 ° W x tan 168 ° 14 ' 

26.3 X tan 358° 16' 

31. 1 — cos -- 2 sin 2 0. 32. tan 2 + cot 2 = 2. 

In the five examples following, find the values of the angles of 
the triangles and expressions for the sides in terms of the angles. 
The triangles are all right-angled. 

33. ab = 12,B=AA. 

Solution : In all cases A -f B — 90°. 

Hence, in this case, 5^4 = 90° : 

.4=18° and B = 72°. 

— = sin A and — = cos A, 
c c 

multiplying — - = sin A cos A, 

c 2 

12 

or by the data — = sin A cos A : 

c 2 



~ am A ona A ' s 



12 



sin A cos A y sin 18° cos 18° 
Also, 

a = c sin A=yj- — P sin ^4=,a/ ]2s1d / = V\2tsmA 

* sin ^4 cos A y cos ^4 



Hence a = Vl2 tan 18° : and, similarly, b = V 12 cot 18°. 

34. a + b == 7, ^4 = f £. 35. — = 3, A = 3jB. 

c 

36. a 2 — 6 2 = 4, #=JU. 37. ?-±-^ 2 = 15, B = \A. 

e 

38. Show that if a-(-^ = f c ? # au( i & cannot be the sides, and 
c the hypothenuse, of a right triangle. 



THE TRIGONOMETRIC FUNCTIONS. 45 

Prove the following equalities : 

39. Biny + co3 y =;=ainysecy + L 

cos ,y 
40 sin_£+_fcan^ = sina . tana . ! - 

cot X -\- CSC X 

4 , tan ft + cot 2 ft _ tan 3 ft + 1 
tan <j> — cot 2 ? tan 3 ft — 1* 

42. sin/3 + cos/? = - 1 + tan/ *. 

sec/3 

43. tana == l^-( 8iDa + cosg -lV 

\ sin a / 

44. a + tanf)(l + oot«-^^ 



45. tan a; — cot x 



Sill cos y 

1 — 2 cos 2 x- 



sin x cos a; 

46. fcana = sinSg , . 

cos a — cos* a 

47. (sin + cos 0) 2 + (sin — cos 0) 2 = 2. 

48. sin (45° ±0) = cos (45° =F 0). 

49. (sin + sec 0) 2 + (cos + esc 0) 2 = (1 + sec esc 0) 2 
5Q tan0 + sec0- l ==tail fl + sec ^ 

tan — sec + 1 

51 tan — sin sec 

sin 3 ~~ l + cos0' 

52. Find the numerical values of 

cos (tan-H), sin (cos _1 |), tan [sin _1 ( — f)], 
tan (cos -1 £) 4- cos (sin -1 *). 
Prove the following equations : 

53. sin- 1 h/3 + tan- 1 h / 3 = ^, |*. -|t, or Mtt. 



54. sin _1 «i = cos _1 V 1 — ra 2 . 

55. sin - J m + cos _1 m = -=-. 

50. tan (cos ~ A) + cot (sin "M.^ = 2 V m 2 — 1. 

57. sin (tan ">) = cos (cot "^). 

58. sin " 1 _4^5_ = tan _1 ( ± 2mn ^ . 

m 2 -j- n 2 \ »ra 2 — n 2 / 



CHAPTEE III. 



GENERAL FORMULAE. 



In Chapter II. have been set forth the relations between 
the trigonometric functions of a single angle. In the present 
chapter will be shown the relations between the functions of 
different angles. 

35. The Sine and Cosine of the Sum and Difference 
of Two Angles. 

Let A OB = any angle, x, and Fig. 17. 

BOC=BOC 1 = any other angle, 
y. We have from the figure 



x + y = AOC, x — y = AOC v 




At D, any point of OB, draw EE X 
perpendicular to OB. Draw EF, 
DG, and E Y H perpendicular to 
OA, and draw DK and E X I paral- 
lel to OA. It is seen at once that 
the triangles OED and EKD are 
equal respectively to OE x D and DIE X ; also, angle KED = 
AOD = x. We have then the following equations : 

FE = GD -f KE = ap _ KE 
OE ' OE OE^ OE' 

HE, _ GD—ID Gp __ KE 
0E X ~ 0E X 
OF = OG — FG 
OE 



sin (x-+y) = 

sin {x — y) 
cos (x -\- y) = 



OE 

OG 



OE' 
KD 



OE 



, , OR OG+GIf 

cos [x — y) = — — = -±— 



OE, OE, 



If, now, we can find values for 



46 



GD 
OE' 



OE 

OG 

OE 

KE 



OE 
KD 

+ OE' 

OG 



(«) 

(fl 

(r) 



OE' OE 1 OE 



GENERAL FORMULAE. 



47 



in terms of the trigonometric functions of x and y, our pur- 
pose will be accomplished. 

Observing that GD and OE are sides of different triangles, 
which have the common side OD, we multiply and divide the 

ratio — — by OD. with th 
OE J 



GD 



GD 
OD 



e following result 
OD 



X 



sin x cos y. 



OE OD OE 

Using ED in the same way, with the next ratio, 
KE 



also, 



and 



OE 



KD 



KE ED 
ED X OE 



OG 



cos DEK sin y 
OD 



cos x sin y ; 



OE 
KD 



0a V - 
OD * OE 



X 



ED 
OE 



COS X COS z/, 
sin DEK sin y = sin x sin ?/. 



OJE ED 

Substituting these values in (a), (£), (p), and (J ), we have 

sin (# -\- y) = sin x cos y -f- cos # sin ?/, (48) 

sin (se — y) = sin # cos ?/ — cos # sin y, (49) 

cos (a? -j- !/) = cos # cos ?/ — sin x sin ?/, (50) 

cos (x — y) = cos x cos ?/ -j- sin x sin ?/. (51) 

These formulae form the basis for deriving almost all the 
others, and for that reason are called " the fundamental for- 
mulae." 

Fig IS 

36. In Fig. 17 both the angles 

x and y were taken, for con- 
venience, less than 90°. The 
formulas, however, are perfectly 
general, and apply to all angles. 
This is shown by the method of 
derivation given below. 

Let AOB = x, and BOC=y 
(any angles). Draw CD perpen- 
dicular to OB (produced if neces- 
sary). Draw CF and DK perpendicular to OA, DG parallel 
to OA, and OE perpendicular to DG. 




48 



PLANE TRIGONOMETRY. 



Then OD = OG cos y, («) 

and DC = OG sin y. (/?) 

Further, since CD is perpendicular to OB, it makes with OA. 
or with its parallel DH, an angle 90° -j- x. 

DG= CD cos (90° + #) 

= — OC sin x sin ?/. by (/?) and (39) 

Also D^=: OD cos ODH= OD cos a? 

= 00 cos a; cos y. by (a) 

And OF=OG cos (# + y). 

Now, taking into account the 
directions in which the several 
lines are measured, 



Fig. 18. 




OF = OK+KF 
= ED + DG. 

Hence, substituting for OF, ED, 
and DG the values found above, 
we have 

00 cos (x -j- y) = OG cos x cos y 
— OG sin x sin y ; 

cos (a? -j- y) = cos a? cos ?/ — sin a? sin y. 

Since in this demonstration no limitation has been imposed 
on the value of y, we may write — y for y. We then have 

cos {x — y) = cos x cos ( — z/) — sin a; sin ( — y), 

or cos (x — y) — cos x cos y -j- sin x sin ?/. by (45) 

If we write 90° — x instead of x in these formulae, we have 

cos (90° — x -\- y) = cos (90° — x) cos y — sin (90° — x) sin y, 
cos (90° — x — y) = cos (90° — x) cos y -f sin (90° — x) sin y, 

which reduce to 

sin (x — ?/) = sin x cos y — cos x sin y, 
sin (a; -j- y) = sin x cos ?/ -j- cos x sin ?/. 



GENERAL FORMULAE. 49 

37. The Tangent of the Sum and Difference of 
Two Angles. 

Dividing (48) by (50), we have, by (23), 

, . x sin x cos y -f- cos x sin y 

tan (x + «) = ^- J —. : — 2- ; 

cos x cos 2/ — sin x sin ?/ 

dividing both terms of the fraction by cos x cos y, 

sin x cos ?/ , cos x sin 2/ 
cos # cos y cos .r cos y 



tan (# -f #) 



cos x cos ?/ sin x sin y 
cos a; cos y cos # cos y 



+ / i \ tan # 4- tan w / RO > 

tan (jj + y) = ^ * . (o2) 

1 — tan x tan y 
In the same manner, by (49) and (51), we have 



sin x cos y — cos x sin y 
from which, as above, 



tan (x — y) 

cos x cos ?/ -j- sin x sin 



tan (.r-y) = *ang-tan y > (53) 

* y 1 -f- tan # tan y v J 

38. Functions of Double Angles. Important formula 
are derived by making y = x in (48), (50), and (52). Thus, 
in (48) we have 

sin (x -j- x) = sin x cos x -j- cos x sin #, 
or sin 2x = 2 sin x cos a;. (54) 

In the same manner, from (50) and (52), 

cos 2x = cos 2 x — sin 2 x, (55) 

tan 2x = 2 tan x . (56) 

1 _ tan 2 x K J 

39. Functions of the Half- Angle. 

Formula (55) expresses the fact that the square of the 
cosine of any angle, diminished by the square of the sine of 
the angle, is equal to the cosine of twice the angle. We may 
therefore write it thus : 

c d 5 



50 



PLANE TRIGONOMETKY. 



cos 2 \x — sin 2 \x = cos x. 
cos 2 \x -\- sin 2 ix= 1. 



Also from (20), 

Subtracting (a) from (/?), 2 sin 2 §x=l — cos x, 

or 



sin 



■—V 1 



COS # 



Adding (a) to (/S) 
or 



2 cos 2 Jx = 1 -f cos x, 



COS *# = / 1 + cos Jg 
Dividing (57) by (58), tan 2 \x 



or 



tan \x 



1 — cos x 

1 -f- cos x 



QO&X 



V 1 -f- cos # 



(a) 

(57) 
(58) 

(59) 



40. Sum and Difference of 
the Sine and Cosine of Two 
Angles. 

Eeferring to Fig. 17, let AOG x 
= and AOC=<p. Then AOB 
= i(f+0) and BOC= IQp — 0). 
We then have 

FE 



Fig. 17 {bis). 
,0 



sm <p 



OE' 



and 



n HE, 

sm 6 = - 



and 

But 
and 

and 
Hence 

and 



OE 1 

sin (p -f- sin = 

sin ^> — sin (9 = 




HE X 
OE' o 

~ OE ~^~ OE ~ OE ' 

i^ff HE, _ FE — HE r 

OE ' 



OE OE 

FE = GD if- jOJ, 

^^ 1= GD — ID = GD — KE, 

FE+HE X = 2GD, 

FE—HE=2KE. 



sin 2> -j- sin = 2 



sin $p — sin 



GD 



OE 
KE 
OE 



2 DG^ GD 
" OB OE' 
2 KE ED 
ED A OE' 



GENERAL FORMULAE. 51 

But Z — = sin A OB = sin I O + 0), 

°P- = cos BOC = cos | (o — 0), 
^- = cos D^ = cos A OB = cos \ U + 0). 

and -— = sin BOC= sin g (cr — 0). 

sin cr -f sin = 2 sin J (e> -f 0) cos | ( cr — 0), 
and sin r — bid = 2 cos I (c -j- 0) Bin I (jp — 0). 

Again, we have 

OF A n OH OH 

cos cr = . and cos = —— = — — , 

(XE 0E 1 OE' 

a OF+OH OH -{-OF 

COS tp — COS = 



and cos <s — cos 6 



OE OE ' 

OF— OH OH— OF 



OE OE 

But 0H= OG + GH = OG + PG, 

and OF=OG — FG. 

0H+ 0F=20G, 
and 0H—0F=2FG. 

Hence cos *, + cos = 2-^ = 2^- X -£?» 

* Q# OB OE' 

and cos w — cos = — 2 — — = — 2 X ^r^r- 

OE BE OE 

But ^^ = cos A OB = cos * O -f 0), 

OB - v- -r y, 

^ = «»2*Oa=coB *(, — *), 

— = — = sin BEK= sin ^ OD = sin h U 
BE BE v ' 

and ^^ = sin B0C= sin | (c? — 6). 



52 PLANE TRIGONOMETRY. 

cos <p -\- cos = 2 cos h (jp -{- &) cos | (<p — 0), 
and cos <p — cos = — 2 sin J (^ -f- 6) sin J (^ — 0). 
This last may also be written thus : 

cos 6 — cos <p = 2 sin h (<p -j- 0) sin J (^> — 0). 

41. The formulse just derived may be obtained analytically, 
without the use of a figure, as follows. Take the sum and 
difference of (48) and (49) and of (50) and (51), thus: 



sin (x -J- y) -j- sin (x — y) = 2 sin # cos 2/, 


(«) 


sin (ic -[- ]/) — sin (x — y) = 2 cos x sin y, 


(« 


cos (x -{- y) -\- cos (# — ?/) = 2 cos x cos ?/, 


(r) 


cos (.r -|" 2/) — cos ( x — V) = — ^ sin # sin ?/. 


(*) 


In these four equations let 




x + y = <p 




x — y = d 




Iding, 2x = <p -\- 0, or x = %(<p -f 0). 





Subtracting 2y = <p — 0, or y = |(^ — 0). 

Substituting these values in (a), (/?), (j), and (5), we have 

sin <p -f sin = 2 sin J (c* -f- 6) cos J (p — 0), (60) 

sin ^ — sin 6 = 2 cos | (^ -f 0) sin J (^ — 0), (61) 

cos ^ -f- cos = 2 cos * (<p -f- 0) cos h (<p — 0), (62) 

cos — cos <p = 2 sin h (y -f 0) sin J (^ — 0). (63) 

If (60) be divided by (61), we have 

sin <p -f- sin sin \ (<p -}- &) cos J (^ — 0) 

sin ^ — sin cos J (<z> -f- 0) sin J (^ — 0) 

= tan J (p -f 0) cot J (^ — 0) : 

sin 9? -f sin tan | (y -|- 0) , fi ,. 

sin <p — sin tan i (ip — 0) 

These formuhe are often useful in putting a given expres- 
sion into a form convenient for logarithmic computation, for, 
it will be observed, they all furnish means of transforming a 
sum or a difference into a product. 



GENERAL FORMULAE. 53 

42. EXERCISES. 

These exercises are introduced at this place to assist the student 
in fixing in his mind the foregoing formulas, to accustom him to 
handling and transforming trigonometric expressions, and to point 
out some of the devices which may be used in such reductions 
and transformations. 

1. To prove (57) by the use of (G3), 

we may write 1 — cos x = cos 0° — cos x. 

By (60) = 2 sin \ (x + 0°) sin \ {x — 0°), 

= 2 sin Ix sin ix, 

= 2 sin 2 2a. 

2. Prove (58) in a similar manner by using (62). 

3. Prove 1 — sin x = 2 sin 2 (45° — ix) = 2 cos 2 (45° + ix). (65) 
Since 1 = sin 2 Ix + cos 2 hx, and, by (54), sin x = 2 sin \x cos \x, we 
may write 

1 — sin x = sin 2 \x -f cos 2 \x — 2 sin \x cos \x, 

= (sin \x — cos ix) 2 . 
Writing sin (90° — \x) for cos \x, this reduces to 

1 — sin x = [sin ix — sin (90° — ix)~\ 2 , 
or, by (61), = [2 cos § (90°) sin i (x — 90°)] 2 

= 4 cos 2 45° sin 2 Q.r — 45°). 
Now cos 2 45° = J, and by (45) sin {%x — 45°) = — sin (45° — }x) ; 
hence sin 2 (ix — 45°) = sin 2 (45° — ix). Making these substitutions 
in the last equation, we have 

1 — sin x = 2 sin 2 (45° — ix). 
Further, since the two angles (45° — ix) and (45° + i%) are together 
equal to 90°, we have 

sin (45° — ix) = cos (45° -f ix). 
Hence we can also write 

1 — sin x = 2 cos 2 (45° -4- ix). 
We may also prove the same thing by writing : 
1 — sin x = sin 90° — sin x, 
by (61) =2 cos i (90° + x) sin J (90° — x\ 

= 2 cos (45° + ix) sin (45° — }x), 
and therefore, since sin (45° — ix) = cos (45° -f- i&), we have 

1 — sin x = 2 cos 2 (45° + ix\ 
and 1 — sin x = 2 sin 2 (45° — ix). 

5* 



54 PLANE TRIGONOMETRY. 

Or, thus : 

1 — sin x = 1 — cos (90° — a;), 
by (57) = 2 sin 2 J (90° — x), 

= 2 sin 2 (45° — %x), 

= 2 cos 2 (45° 4- %x), as before. 

4. Prove 

1 + sin x = 2 sin 2 (45° + \x) = 2 cos 2 (45° — £z) (66) 

in each of the three ways given above. 

5. Develop the formula for sin (x -\-y -\-z). 

We write first sin [x 4- y 4- 2) = sin [(a; 4- ?/) + 2]. 
If this be developed by (48), we have 

sin {x-\-y-\-z) = sin (a; 4- y) cos 2 4- cos (x 4- 2/) sin 2, 
then substituting the values of sin (x 4- y) and cos (x 4- 2/), we have 
finally 

sin (a; 4- y 4- 2) == sin a; cos 2/ cos 2 4- cos x sin ?/ cos 2 
4- cos x cos y sin 2 — sin x sin y sin z. 

6. In the same way prove the following formulae : 

(1) sin (x 4- y — z) = sin # cos ?/ cos 2; 4- cos a; sin 7/ cos z 

— cos x cos y sin 2 4- sin x sin ?/ sin z. 

(2) sin (# — y — 2) = sin a; cos ^/ cos — cos x sin 3/ cos z 

— cos x cos y sin 2 — sin x sin ?/ sin z. 

(3) cos (a; 4- y 4- 2) = cos a; cos y cos 2 — sin x sin 2/ cos 2 

— sin a; cos y sin 2 — cos x sin ?/ sin 2. 

(4) cos (x -\- y — z) = cos x cos ?/ cos 2 — sin a; sin y cos 2 

4- sin x cos ?/ sin 2 4- cos x sin 2/ sin 2. 

(5) cos (a; — y — 2) = cos a; cos 2/ cos 2 4- sin a; sin y cos 2 

4- sin x cos 2/ sin 2 — cos x sin ?/ sin 2. 

7. Show that 

tan (x 4- 2/ + 2) = tan a? 4- tan ?/ 4- tan 2 — tan a; tan ,y tan 2 
1 — tan x tan ?/ — tan y tan 2 — tan 2 tan a/ 

8. To derive the formula for sin 3a; we put x=y — z in the 
formula of Exercise 5. Thus we find after reduction 

sin 3a; = 3 sin x cos 2 x — sin 3 x, 
substituting cos 2 x = 1 — sin 2 x, we have 

sin 3a; = 3 sin x (1 — sin 2 x) — sin 3 x ; 

sin 3a; — - 3 sin x — 4 sin 3 x. (67) 

9. In a similar manner show that 

cos 3a; = 4 cos 3 x — 3 cos x. (68) 

10. Deduce a formula for tan 3a; from that given in Exercise 7. 



GENERAL FORMULAE. 



55 



43. The following formulae, which are of frequent use, are 
left for the student to demonstrate : 



cos — cos (p 



tan J O -f 0) tan \ (<p — 0), 



cos + cos <p 

sin (<p + 0) tan <p + tan 

sin (<p — 0) tan <p — tan (J 

cos Q + 0) _ 1 



tap <p tan 



cos (<p — 0) 1 + tan 9? tan 0' 

tan(0 + 45°) = 1 + ta ^-g 
^ ; 1 — tan 



tan (0 — 45°) 



tan — 1 
tan 0+1* 



(69) 

(70) 
(71) 
(72) 
(73) 



44. Small Angles and Arcs. If the angle ACP, or 0, 
is very small, we may make cer- 
tain assumptions with regard to 
it which are not true of angles in 
general. We have, Fig. 19, 

AP 

— - = (in radians), 



tan 0. 




MP . a MP 

= sin 0, 

CP ' CM 

As approaches the limit 0, CM 
approaches CA or CP, as a limit. 

TT MP , iliP 

Hence — — : approaches -— — as a 

limit. Upon the same supposition the arc AP and the 

straight line MP approach each other in value. Therefore 

AP MP 

— — approaches — — . Hence as approaches the limit 0, 

L/Jl (yJr 

0, sin 0, and tan approach equality. We may therefore write 
for very small angles, within a certain limit of error, 

= sin = tan 0. (74) 

The limit of error within which this equation is true 
diminishes with the angle, but it must be remembered that 



56 PLANE TRIGONOMETKY. 

there is always some error. A table of natural functions 
shows that for angles up to and including 30' the equation is 
true as far as the sixth place of decimals. Thus, 

30' = .008727 == sin 30' == tan 30'. 
The same is true to five places of decimals for angles up to 1°. 

If d be such an angle that md <[ 1°, we have, to five decimal 

places, by (74), 

md = sin md — tan md ; 
also, by multiplying (74) by m, 

md = m sin d = m tan : 
.-. within the limits specified, 

sin md = m sin d, 
tan md = m tan d. 

EXAMPLES. 

1. Determine the functions of 15° and 75° from those of 45° and 
30°. Use (48) and (50). 

2. Determine the functions of 22 }° from those of 45°. Use (57), 
(58), and (59). 

3. Determine the functions of 120° from those of 60°. Use (54), 
(55), and (56). 

4. Determine the functions of 0° from (49), (51), and (53), by 
making y = x. 

5. From the results of Ex. 4 find the functions of 90°. 

6. From the results of Ex. 5 find the functions of 180°, 270°, and 
360°, using (48) and (50). 

7. Verify equations (39), (40), (41), (42), (43), (44), and (45), by 
making suitable substitutions in (48), (49), (50), (51), (52), and (53). 

Prove the following relations, as far as Ex. 32 : 

8. tan 6 + cot = 2 esc 20. 

9. Bina; + 8in ^ = tan*(^ + y). 

cos x + cos y 

10. ring - Bin y = tanHs--y). 

cos x -f- cos y 

1L su L x±^ SL y == _ cotUx _ yy 

cos x — cos y 
12> sin x - sin y = _ cot , {x + }> 

cos x — cos y 



GENERAL FORMULAE. 57 

13. cot (* ± y) = wt«eot»=Fl . 14 . ta „ ix = !-«» , 
cot 2/ db cot x sin .»■ 

15. tauk^i- j; 8 !'^- 00 " - 16. W« = l-*"',% 

1 + sm a; + cos a; 1 + tan- > 

<*, * n-o , ia\ Jl + sin0 14-sinfl _ cos 

17. tan (4o°4- 2 0) =\— ^ — : — ~ = — a — ; : — «" 

v ' ' 1 — sin cos 1 — sin 

18. cos == 2 cos (45° + i*) cos (45° — £0). 

2 

19. cos a 



tan (45° + J0) + (tan 45° — £0) 

20. sin (a: + g/) sin (a; — y) ■== sin' 2 x — sin 2 y = cos 2 y — cos 2 x. 

2 1. cos (x + y) cos (a: — y) — cos 2 a; — sin 2 y = cos 2 # — sin 2 x. 

22. cos 6 ± sin = ^2 sin (45° ± 0) = >/2 cos (45° =f 0). 

23. tama±tani8 = Hin ^ ± ^ . 

cos a cos j3 

24. cota±cot/3^ sin ( a=b ^ 

sin a sin p 

25. 2 sin 2a z= (sin a -|- cos a) 2 — (sin a — cos a) 2 . 

26. esc 20 = i sec esc 0. 

27. sin 2<j> + cos 2^ = (sin <j> + cos <£) 2 — 2 sin 2 0. 

28 1 + 2 siu a = tan ^ 15 ° + - a ^ 
1 — 2 sin a tan (15° — £a)" 

29. (cos 90° — cos 2x) — (cos 90° + cos 2a;) = 2 (sin 4 x — cos 4 a'). 

30. sin oa -(- sin 3a = 2 sin 4a cos a. 

3 1. 2 esc 40 4- 2 cot 40 = cot — tan 0. 

32. g° 9g + siDa = sec2« + tan2a. 

cos a — sin a 

33. Given sin = -J. Find all the functions of 20 and Id. 

34. Given cos = f. Solve in the same manner. 

35. If tan a = \ and tan p = £, what is tan (a 4- 23) ? 

36. Show that each of the following expressions is equal to sin x : 



(1) 

(2) 


cos (30° — x) — 

2 tan \x 
1 + tan 2 \x 

1 
tan \x 4- cot x 

1 — 2 sin 2 (45° - 


cos (30° 
-ix). 


4- x). 

(3) 

(5) 

(<) 


2 
cot \x 4- tan \x 


(4) 
(6) 


1 tan a; 
* 2 (tan a; -f cot 2a-)' 

Vl 4- sin 2a:— Vl— sin 2.u 



58 PLANE TRIGONOMETRY. 



37. Show that the following expressions are equal to cos : 

(1) I . (2) 1 + cos2 ^ 

w 1 + tan 6 tan £0 w 2 cos ' 



,„, Vl + sin2d+Vl — sin 20 
(3) g 

38. Show that the following expressions are equal to tan a : 
m sin 2a ^) 1 — cos 2« 



1 -4- cos 2a sin 2a 

2 ,,n 1 -j- tan a cot 2c 

cot £a — tan %a cot a 4- cot 2a 



(5) 
(6) 
(7) 



V / l-|-sin2a — cos a 
l/l + sin 2a — sin a 
sin la cos \a 



cos (45° + £a) cos (45° — \a) 

tan (45° + \a) — tan (45° — \a) 
2 

Determine all values of < 360° in the following : 
39. tan 20 = 3 tan 0. 40. tan 6 = tan 40. 

4 1. 3 tan tan 30 + 1 = 0. 42. sin = tan — tan 20. 

43. sin + sin 20 + sin 30 = 1 4- cos + cos 20. 

44. Prove that if sin 5a — sin 3a = tan 2a — tan a 1 
then sin 8a = 4 sin a. 

45. Find the values of the following expressions : 

(I) sin (sin -H + sin- 1 1/2). (2) cos (sin- 1 ! — tan- 1 !). 
(3) sec (tan- 1 ! -f cos- 1 £). (4) tan (sin- 1 ! — cos- 1 !). 

(5) sin (sin -1 ?™ -4- cos -1 ??). 

(6) tan (cos- 1 — sin- 1 -^. 

V V i/a j/a/ 

46. Find the values of the following expressions : 

(1) tan (2 tan ~ 1 2). (2) sin (J cos- 1 !). 

(3) tan (J cos-H). (4) cos (2 tan- 1 \y 3). 

(5) sin (3 tan "V2). (6) cos (3 sin- 1 jys). 

(7) sec f2cos- i m ~~ w Y (8) tan U sin- 1 ^^. 

(9) cos (Jtan-^/3 — 2sin-H T / 2 )- 

47. Show that 

tan - 1 (cos 20) = | tan - 1 [J (cot 2 — tan 2 0)]. 

48. Show that 

sin ftan- 1 2 tan *» 1 = 2 Bi " *' - 
L 1 — tan 2 ^J 1/1 + tan 2 ^ 



GENERAL FORMULAE. 59 

45. Trigonometric Tables. A trigonometric table con- 
sists generally of two parts, — one which gives the trigo- 
nometric functions of each angle, and the other which gives 
the logarithms of those functions. The first is called a table 
of natural sines, cosines, etc., and the second a table of loga- 
rithmic sines, cosines, etc. As most of the computations of 
trigonometry are carried on by means of logarithms, the 
latter is used more than the former. 

The functions themselves, and also their logarithms, being 
incommensurable numbers, can be obtained only with more 
or less approximation by employing a greater or less number 
of decimal places. We find tables of logarithmic functions 
computed to four, five, six, seven, ten, or more places. The 
table to use in any case is to be determined by the character 
of the work and the degree of precision required in the 
result. Four-place tables generally give sine, cosine, tangent, 
and cotangent for every ten minutes of arc, with differences 
and proportional parts for interpolating minutes. Five-place 
and six-place tables generally give the functions for every 
minute, although sometimes in the latter the functions are 
given for every ten seconds. In some five-place and in all 
six-place tables proportional parts are given for interpolating 
seconds. Other five-place tables give proportional parts for 
interpolating tenths of a minute only. Seconds cannot be 
determined accurately in all parts of the quadrant with a five- 
place table. Seven-place tables give the functions for every 
second, and tenths, and in some cases hundredths, of a second 
can be interpolated. No tables give directly the functions 
of angles beyond the first quadrant. It has already been 
shown how the functions of other angles may be found from 
these. See §§ 28-32. 

46. If the student will examine, by means of Fig. 16, the 
variation of the sine as the angle increases from 0° to 90°, he 
will see that it increases much more slowly when the angle is 
near 90° than when it is near 0°. This can also be seen by 
examining a table of trigonometric functions. When the 
angle is near 90° the log sin will be found to change very 



60 PLANE TRIGONOMETRY. 

slowly, often several successive logs being alike, while near 0° 
it changes very rapidly. It follows, therefore, that if we are 
seeking to determine an angle which is near 90°, we should 
endeavor to avoid finding it from its sine, as in that case our 
determination will lack precision. Since the values of the 
cosine are the same as those of the sine in the reverse order, 
it follows that we should never determine a small angle by its 
cosine. 

The variation of the tangent is rapid at all times ; it may, 
therefore, be used equally well at all parts of the quadrant. 

A detailed description of the method of using the tables 
is not necessary here, as it generally accompanies the tables 
themselves. 



CHAP TEE IV. 

SOLUTION OF RIGHT TRIANGLES. 

47. To Solve a Triangle means to find the values of 
the unknown parts from the given values of the known parts. 
Any triangle can be solved provided three parts, one of which 
is a side, are given. Since one angle of a right triangle is 
always 90°, it follows that a right triangle can be solved if 
either a side and an acute angle, or two sides, are known. 

48. Ordinary Method of 
Solution of Right Triangles. 
The formulas for solving right 
triangles are derived at once by 
applying definitions of the trigo- 
nometric functions, § 12, to Fig. 5. 
Thus we have : 



Fig. 5 {bis). 




sin A 



cos A 



tan A 



cot A 



a 

> 


sin B = 


b 


c 




c 


b 


cos B = 


a 


c 




c 


a 


tan B = 


b 
a 


b 
a 


cot B = 


a 

T 



(5) 
(6) 
(7) 
(8) 



For convenience these formulas are numbered as in § 12. 

The way in which the solution is performed can best be 
learned by a study of the following examples : 

1. Given c = 72.98, 5 = 19° 37'. 

In every right triangle ^4 — i? = 90°. Hence, if one angle is 
given, the other can be found immediately by subtracting the 



Thus, 



90 c 



19° 31 



ro°2c 



61 



62 PL AXE TRIGONOMETRY. 

To find the two sides a and b, we look at the formulae above 
and see which of them will furnish equations in which a and b 
are the only unknown quantities. In this case either (5) or (6) 
will answer. Let us take 

sin B = — and cos B = —. 

c c 

This selection is made, so that both a and b may be found from 
the two parts c and B, originally given. From these two equa- 
tions, therefore, we have 

b = c sin B and a = c cos B. 

log c = 1.86320 log c = 1.86320 

log sin B = 9.52598 log cos B = 9.97403 



log b = 1.38918 log a = 1.83723 

6 = 2-4.50 a = 68.74 

If the work has been correct, we shall have 



a = Vc 2 — b 2 = V{c — b) (c + b) 

c — b = 48.48 log = 1.68556 

e + b = 97.48 log = 1.98892 

log a 2 = 3.67448 
and log a = 1.83724 . •. a = 68.74 

The last operation is called the u check." 
2. Given b = 93.68, A = 41° 6' 12". 
First, B = 90° — A = 48° 53 / 48". 

Selecting the formulae as before, we have 

tan A = — and cos A = — , 

b c 

or a = b tan A and c 



cos A 

log b =1.97165 log b =1.97165 

log tan A = 9.94074 log cos A = 9.87710 



= 1.91239 logc =2.09455 

a = 81.73 c = 124.32 



Check formula b=v(c — a) (c + a). 

G — a= 42.59 log =1.62931 

c+a = 206.05 log = 2.31397 

log b 2 = 3.94328 
and log b =1.97164 .-. 6 = 93.68 



SOLUTION OF RIGHT TRIANGLES. 63 



3. Given a = 1.9436, c = 3.0074. 
Here the formulae are 

sin A = — , b = c cos A, and B = 90° — X 
c 

log a = .28861 log c = .47819 

We = .47819 log cos A = 9.88259 



log sin A = 9.81042 log 6 == .36078 

A = 40° 15' 40" b = 2.2950 

B = 90° — A = 49° 44' 20". 



Check formula 6 = v (c — a) (c + a). 

c — a = 1.0638 log = .02686 

c-{-a = 4.9510 log = .69469 



log b 2 = .72155 
mid log b = .36078 . • . 6 = 2.2950 

4. Given a = 639.41, b = 417.93. 
Here the formulae are 

tan -4 = -?-, c = -t^-, and^ = 90° — ^. 
6 sin A 

log a = 2.80578 log a =2.80578 

log 6 = 2.62110 log sin A = 9.92276 

log tan A = 10.18468 log c = 2.88302 

A = 56° 49' 51". c = 763.87 

.B = 90° — .4 == 33° 10' 9". 



Check formula a = v(e — b) (c + 6). 

c — 6= 345.94 log =2.53900 

c + 6 = 1181.80 log = 3.07255 

log a 2 = 5.61155 
and log a =2.80578 .-.a = 639.41 

49. In working problems of this kind, where certain quan- 
tities are to be determined from other given quantities by a 
more or less extended arithmetical process, care should be 
taken to compute each of the unknown quantities, as far as 
possible, directly from the data. If this is done, an error 
made in computing one part of the result is not carried on to 
vitiate the other parts. In Ex. 3, above, this principle was 
disregarded, since A was used in finding b. This was done so 
that the formula b = V (c — a) (c -f- a) might be reserved for 



64 PLANE TRIGONOMETRY. 

the check. In Ex. 4 we are obliged to use A in determining c. 
It is true that c 2 = a 2 -j- b 2 is theoretically the proper formula 
to use here ; but it is not convenient for logarithmic computa- 
tions, in that it involves the sum of two squares. 

In constructing the check formula for these cases, and for 
all others where such a formula is used, it is essential that it 
should contain none of the quantities used in the direct com- 
putation. For it is only by thus making the " check" an 
entirely independent operation that it can be of any real 
value in testing the accuracy of the result. 

EXAMPLES. 
Solve the following right triangles : 

1. A = 72° 13', c = 429.3. 9. A = 40° 41' 24", 6 = 729.87. 

2.^ = 48° 16', a = 37.298. 10. a = .4723, 6 = 1.0000. 

3. a = 412.96, 6 = 278.41. 11. c = 1.002, a = .492. 

4. A = 36° 42' 18", a = 412.13. 12.^ = 34°, c = 5. 

5. e = 4792.3, 6 = 3874.7. 13. 6 = 47, c = 72. 

6. e = 38.961, ^4 = 29° 13' 12". 14. a = 4.002, A = 61° 21'. 

7. a = 1.2367, 6 = 2.0072. 15. a = 12.606, B = 58° 3'. 

8. B = 41° 9' 18", 6 = 14.142. 

50. Solution of Right Triangles when a Side and 
the Hypothenuse are nearly Equal. When one side 
and the hypothenuse are given, and these given parts are 
nearly equal in value, the above method of solution will be 
found undesirable. Thus, let b and c be the given parts, and 
let them differ but little in value. Formulae (5) and (6) give us 

sin B = — and cos A = — . 
c c 

l^ow, from the nature of the case, B must be nearly 90° and 
A must be very small, so that both of these formulae are un- 
desirable. (See § 46.) To overcome this objection we make 
use of the following method. 

From c 2 = a 2 -f- b 2 

we have a 2 = c 2 — 6 2 , 

or a = V{c — 6) (c -f 6), (75) 

by. which a can be found. We then have 



SOLUTION OF EIGHT TRIANGLES. 65 

sin A = — and B = 90° — A. 
c 

We can use for our check formula 

Bin B = — , 
c 

which is sufficiently accurate for that purpose. 

Example: Given c = 147.92, 6 = 146.39. 

log (c — b) = .18469 



c = 


147.92 


6 = 


146.39 


c — 6 = 


1.53 


c + 6 = 


294.31 


a = 


21.22 


^4 =8° 


14' 52" 


,B =81 


D 45' 8" 


B =81 


D 45' (check). 



log (c + 6) 


= 2.46881 


log a 2 


= 2.65350 


log a 


= 1.32675 


logo 


= 2.17003 


log sin A 


= 9.15672 


log 6 


= 2.16551 


logc 


= 2.17003 


log sin B 


= 9.99548 



The student can see from this that if the above solution had 
been performed by the method of § 48 the value found for B 
would have been that found by the check formula, and hence 
both the angles would have been 8" in error. 

51. Another method of solving a right triangle in which 
c and b are the known parts and are nearly equal is the 
following: 



Since 


A V 

cos A = — , 
c 


then 


1 — cos A = : 

c 


.-.by (57), 


2 sin 2 i A — C ~~ b , 
c 


or 


Bini^-^/ 2c • 


Then 


a = c sin A and B = 90 — A ; 


and for a check 


formula, b = \/(c — a) (c -j- a). 



(76) 



(If c and a are given, use a = y(c — b) (c + 6) for the check 
formula.) 



66 PLANE TRIGONOMETRY. 

This method is especially useful when only the angles are 
required, as (76) gives the result directly from the data, with- 
out first finding the unknown side. 

Example : Given c = 117.92, 6 = 146.39. 

c = 147.92 log (c — b) = .18469 



b = 146.39 


log 2c = 2.47106 


c — b= 1.53 


log sin 2 \A = 7.71363 


2c = 295.84 


log sin \A = 8.85681 


\A = 4° T 26" 




A =8° 14' 52" 


log c = 2.17003 


B = 81° 45' 8" 


log sin A = 9.15671 


a = 21.22 


log a = 1.32674 




EXAMPLES. 



Solve the following right triangles : 

1. a = 42, c = 43. 4. c = 1000, a = 998.2. 

2. 6=1.0092, c = 1.0162. 5. a =50.972, c = 60. 

3. a = 402.96, c = 403.17. 

52. Area of a Right Triangle. Let the student prove 
the following formulae, where k represents the area of a right 
triangle : 

A = lab. (77) 

k = lc 2 sin 2A = \c 2 sin 2B. (78) 

(Note. — To prove (78) substitute in (77) a = c sin A, b == 
c cos A ; or a = c cos B, b = c sin 5.) 



A- = Ja i/(c — a) (c -f a) = £6 l/(c — 6) (c + 6). (79) 



(Note. — To prove (79) remember that a = y(c — -b) (c -J- 6) 
and b = V (c — a) (c -J- «).) 

A = *a 2 cot 4 = £& 2 cot B. (80) 

A = lb 2 tan A = Aa 2 tan B. (81) 

EXAMPLES. 
Solve the following right triangles : 

1. 6=2.13, c = 298.61. 3. a = 522.1, 6 = 523.2. 

2. a = 1.02, 6 = 192.16. 4. & = 4296, ^4 = 36° 12'. 



SOLUTION OF RIGHT TRIANGLES. 67 

Solution of Example 4. Formulae (SO), (81), and (78) give us 



* t;in -4 * sin ' 



>to 6= ^to e=2 Atai' 



by which we can solve. a = V (c — 6) (c -f- 6) will be an efficient 
check formula. 

5. k = 12.93, a = 2. 7. k =-3.924, i? = 72 12'. 

6. A; = 368.9, c = 126.12. 8. £ = 2116, p = 12. 

Solution of Example 8. In this and the following examples p 
means the perpendicular from the right angle to the hypothenuse. 
The formulae for solving this example are obtained as follows : 

Since %pc = k, Ave have c — -. (a) 

P 

Then, from (78), sin 2.4 = -±f. 

If this be combined with (a), eliminating c, we have for sin 2^4, 
expressed directly in terms of p and k, 

sin 2A = *&-. (i3) 

k 

With c and A known, the rest of the solution may be performed 
as in | 48. It is possible to express a and b directly in terms of k 
and p, but the formulae are not convenient for logarithmic compu- 
tation. For a check formula use 6 = V {c — a) (c + a). 

In the next seven examples give formulae for the general solu- 
tion as in Exs. 4 and 8, above, before performing the logarithmic 
work. 

9. c= 92.68, p = 26.92. 13. p = 161.2, 5 = 43° &. 

10. £ = 9368, a + 6 = 738. 14. p = 3.92, a — 6 = 1.41. 

11. « + 6 = 16.92, c = 12.13. 15. £ = 262.98, B — A = 1Q°12'. 

12. a + b = 621.31, a — 6 = 103.17. 

Prove the following relations between the parts of a right 
triangle, Exs. 16-24 : 

16. sin 2A = sin 2B. 20. cos 2^1 = sin {B — A). 

17. a 2 cot A = b 2 cot B. 21. cos2^ = (6 + a) (& — «) 

c 2 

18. tan^=^=^. 22. cos(^ — .4)=^. 

a ; c 2 

19. tan (45° ± A) = A±^ 2 3. tan (B - A) = {b + q > ( 6 ~ a) 

; 5 + a v } 2ab 



68 PLANE TRIGONOMETRY. 

24. abc = a 3 cos A + 6 3 cos i?. 

25. Prove that in every right triangle the value of c lies between 

•^J±J>. and a + 6. 

t/2 

26. In a circle of radius 1, given the angle at the centre 0, to 
determine the chord subtending the angle. 

(1) 6 = 43° 2'. (2) = 75° 10' 24". (3) = 125° 1& 36". 

27. In a circle, given the radius r, and the angle at the centre 0, 
to determine the chord. 

(1) = 27° 42' 36", r = 4.176. (2) = 168° 6', r = 14.47. 

28. In a circle of radius 1, given the length of a chord s, to 
determine the angle at the centre which it subtends. 

(1) s = 0.74922. (2) s = 1.76588. 

29. In a circle, given the radius r = 7.6510, and a chord s = 
8.4458, to determine the angle at the centre subtended by the 
chord. 

30. In a regular polygon of n sides, inscribed in a circle, de- 
termine the lengths of the side and apothem, if the radius of the 
circle is unity. 

(1) n = 4:. (3) n = 6. (5) n = 8. (7) » = 10. 

(2) n = 5. (4) n = 7. (6) n = 9. (8) n = ll. 

31. How great is the angle at the centre of a circle, if the chord 
subtending the angle is two-thirds of the radius ? 

32. Determine the area of the segment of a circle, having given 
the radius r = 11.2840 and the chord s = 6.9738. (Note that there 
will be two solutions.) 

33. From a point in the circumference of a circle of radius r = 5, 
two chords are drawn, whose lengths are respectively a = 2.860 
and 6 = 7.098. Determine the angle between the chords. (Note 
that the figure can be constructed in two ways.) 

34. In a circle of radius r = 57.294, two parallel chords are 
drawn, their lengths being respectively a = 105.48 and 6 = 111.65. 
Determine (1) the arcs intercepted between these chords, and (2) 
the portion of the area of the circle that lies between them. 

35. The area of a regular heptagon is jP= 43.253. Determine 
the length of its side. 

36. The side of a regular dodecagon is 11.2 feet. Determine the 
radii of the inscribed and circumscribed circles. 

37. A regular enneagon whose area is 289.25 square metres is 
inscribed in a circle. What is the radius of the circle? 



SOLUTION OF RIGHT TRIANGLES. 60 

38. The area of a regular pentagon inscribed in a circle is 3.3184 
square centimetres. What is the area of a regular polygon of 
eleven sides inscribed in the same circle ? 

39. Regular polygons of eleven and twenty sides have equal 
perimeters; and the area of the latter exceeds that of the former 
by 294.56 square metres. How long are the sides of the polygons, 
respectively? 

40. The area of a parallelogram is 668.11 square feet, one side is 
13.5 feet and one angle 36° 36' 10". Determine the other sides. 

41. One angle of a rhombus is 64°, and the shorter diagonal is 
123 feet. Find the lengths of the sides and of the other diagonal. 

42. The middle points of the sides of a rectangle form the 
vertices of a rhombus. Determine the angles of the latter from 
the sides a = 1.3782 and b = 4.S063 of the former. 

43. Determine the angle which two tangents to a circle make 
with each other, having given the radius = 1.67, and the distance 
from the centre of the circle to the point of intersection of the 
tangents = 4.66. 

44. How far must a person be elevated above the surface of the 
earth in order to be able to see an object on the surface at a dis- 
tance of 15 miles? Take the radius of the earth as 4000 miles. 

45. Determine the angle between the outer common tangents 
to two circles, having given the radii of the circles, B = 6.130, 
r = 2.014, and the distance between the centres = 8.49. 

46. In what ratio will the area of an isosceles right triangle be 
divided by the bisector of one of the acute angles? 

47. A right triangle is divided into two equal parts by a straight 
line drawn from one of the extremities of the hypothenuse. If 
the angle from whose vertex this line is drawn is 60°, in what 
ratio will it be divided? 

48. In an isosceles trapezoid the area and parallel sides are 
respectively 3.4771, 5, and 3. Determine the angles. 

49. The non-parallel sides of a trapezoid are respectively 3.51 
and 7.04 metres, and if produced they will intersect at right angles. 
Determine the angles of the trapezoid. 

50. Determine the angle of intersection of two circles whose 
radii are R = 527.39, r = 474.27, and whose common chord is a = 
422.00. 

51. In a right triangle in which a > 6, c — a = a — b. Deter- 
mine the angles. 




70 PLANE TRIGONOMETRY. 

52. In an isosceles triangle the angle at the vertex is A, and 
the perpendicular from one extremity of the base to the opposite 
side is h. Determine the area. 

53. Upon the top of a cliff stands a post 5 feet high. At a hori- 
zontal distance of 400 feet from the base of the cliff, the top of 
the post has an angle of elevation of 61° 2'. Determine the height 
of the cliff above the observer's eye. 

54. From the top of a light-house 110 feet above the level of the 
sea the angle of depression of a ship is 4° 14'. Determine the 
distance of the ship from the foot of 

the light-house. Fig. 20. 

55. To determine the height of an 
object, B, above a level plane, AM, 
two stations, M and JV, are selected, 
the distance between them is meas- 
ured, and the angle of elevation of 
the object at each station is meas- 
ured. From these data the elevation 

can be computed. If MN=d, BMA = d y and BNA — (p, and h 
be the required height, show that 

t d sin <f> sin d 

~~ sin {(p — 0) ' 

56. If d = 329.86 feet, tf> = 59° 12', and = 41° 21', find h, and 
also the distance MA. 

57. Upon the top of a tower 125 feet high stands a pole which, 
at a horizontal distance of 200 feet from the foot of the tower, 
subtends an angle of 49'. How long is the pole ? 

58. A balloon, whose diameter is 2 metres, ascends vertically 
from a point A. After it has risen a certain distance an observer 
at a horizontal distance of 800 metres from A observes that the 
balloon subtends an angle of 4' 30". How high is it when the 
observation is taken ? 

59. At A the angle of elevation of a cloud in the S.W. is 43° 
35' 5". At B, 2526 metres due south from A, the cloud appears in 
the N.W. What is its angle of elevation at B, and what is its 
distance from the earth ? 

60. Upon a hill overlooking the sea stands a tower 65 feet high. 
From a ship the angles of elevation of the foot and top of the 
tower are respectively 14° 2' and 14° 34'. What is the horizontal 
distance of the ship from the tower, and what is the height of the 
hill? 



SOLUTION OP RIGHT TRIANGLES. 71 

61. A tower and a monument stand on the same horizontal 
plane. The height of the tower is 29.3 metres, aud the angles of 
depression of the top and foot of the monument from the top of 
the tower are respectively 3° 45' 42" and 4° 11' 22". How high is 
the monument? 

62. From the top of a tower 100 feet high the angles of depres- 
sion of two objects on the plane upon which the tower stands are 
respectively 17° 39' and 29° 16'. Find the distance between the 
objects (1) when they are on the same side of the tower, (2) when 
they are on opposite sides of the tower, (3) when the lines joining 
them with the foot of the tower are at right angles. 

63. Two towers stand on a horizontal plane at a distance of 120 
feet from each other. A person standing successively at their 
bases observes that the angle of elevation of one is twice that of 
the other, but when he is half-way between them their angles 
of elevation are complementary to each other. What are the 
heights of the towers ? 

64. A regular pyramid has for its base a square whose side is 
458.2 feet. The angle between any one of its faces and the hori- 
zontal plane is 59° 36'. Find the angle its edges make with the 
plane, its slant height, the length of an edge, and the altitude of 
its apex. 

G5. At the junction of two roads, AB and AC, whose angle is 
55°, a surveyor wishes to lay off a triangular lot, containing 87,120 
square feet, by a line at right angles to AB. What will be the 
frontage of the lot on each road? 

If the lot is to be in the shape of an isosceles triangle with its 
apex at the junction of the roads, what frontage will it have on 
the roads ? 

If the lot is to be in the form of a quadrilateral cut off by equal 
perpendiculars to the two roads, what will be its frontage? 

Which lot will be the least expensive to fence ? 

QQ. The length of a degree of longitude on the equator is ap- 
proximately 69.1 miles. Determine the length of a degree of 
longitude at Philadelphia, 40° N. latitude. 



CHAPTER Y. 

FORMULAE FOR THE SOLUTION OF OBLIQUE TRIANGLES. 

53. In any triangle the sides are proportional to the sines of 
their opposite angles. 

Let ABC be any triangle. We Fig. 21. 

are to prove 

a b c 

sin A sin B sin C 

Let fall a perpendicular from 
C to AB. We have then in both 
figures 



sin A 



V 



or p = b sin A, 



and sin B = I—, or p = a sin B: 
a 

a sin B = b sin A, 

a b 

sin A sin B 

In a similar way we can show 

b c -x c 




or 



sin B sin C 
a 



and 



a 



sin C sin A 
c 



(82) 
sin JL sin B sin 

54. in any triangle the sum of two sides is to their difference 
as the tangent of half the sum of the opposite angles is to the 
tangent of half their difference. 

From (82), — = — — — . 

v ; ' b sinJ5 

72 



(83) 



SOLUTION OF OBLIQUE TRIANGLES. 73 

By composition and division, 

a -j- b sin A -f- sin B 

a — b sin A — sin B 

-d , , /n , N sin A -\- sin B tan $ (A -J- B) 

g-f- 6 _ tan £ (A -f- ff) 

a — 6 _ tan £ (A — B)' 

with similar expressions for the other pairs of sides. 

35. In any triangle the square of any side is equal to the sum 
of the squares of the other two sides minus twice the continued 
product of these two sides and the cosine of the included angle. 

We are to prove 

a 2 = b 2 -f c 2 — 2bc cos A. 

In the first triangle, Fig. 21, 

BD = AB — AD, 

and in the second, 

BD = AD — AB; 
but in both cases 

BD 2 = AB 2 + AD 2 — 2AB . AD. 
To both sides of this equation add CD 2 , then 

BD 2 + CD 2 = AB 2 + AD 2 + CD 2 — 2AB . AD ; 
but BD 2 + CD 2 = BC 2 , 

AD 2 -f CD 2 = AC 2 , 
and AD = AC . cos A : 

^C 2 = AB 2 + ZC 2 — 2.45 . ZC. cos .A, 
or a 2 = b 2 -f- c 2 — 26c cos J.. 

In both of these cases the side a is opposite an acute angle. 
To prove the same thing for a side opposite to an obtuse 
angle, we have, in the second triangle, Fig. 21, 

AD = AB + BD : 

AD 2 = AB 2 -\- BD 2 -f 2AB . BD. 



74 



PLANE TRIGONOMETRY. 



Adding CD 2 to both sides, we Fig. 21. 

have 

ZD 2 + CD 2 = AB 2 -f BB 2 + CD 2 

-j- 2A5 . BD. 
But ZD 2 + CD 2 = b 2 , 
BD> J r (W 2 = a 2 , 
and 4 BD = a cos BBC 
= a cos (w — E) = — a cos B ; 
.-. 6 2 = c 2 + a 2 — 2ca cos 5. 
Hence we may write in any tri- 
angle, 

a 2 = b 2 -f c 2 — 26c cos ^; 
fr 2 — c 2 -j- ^ 2 — % ca cos i?, 
c 2 =a 2 +b 2 — 2abcos C) 

53. To determine the sines of the half-angles of a triangle in 
terms of the sides. 

From the first of (84) we have 




(84) 



cos A 



b 2 -\- & 



2bc 



Subtracting both sides of this equation from unity, we have 



cos A 



2bc 
__2bc—b 2 —c 2 +a 2 
2bc 

2bc 





(a-f5-c)(fl-Hc). 
2bc 


.-.by (57), 


• 2 i a (a -\-b — c) (a — b 4- c) 
sin 2 \A = * — L_ LA ! — /. 

4bc 


Now, let 
Subtracting 


a -j- 6 -f- c = 2s 
2c = 2c 



Likewise, 



a -j- b — c = 2 (s — c). 
a — 6 -f c = 2 (s — 6). 



SOLUTION OF OBLIQUE TRIANGLES. 



75 



also, 



• 2 i a (s — b)(s — c) 
sin 2 h A = * ^ J - 

bc 

(s — c)(s — <Q 
ca 

(s — a)(s — b) 



sin 2 $B 

sin 2 W 



ab 



(85) 



57. To determine the cosines of the half-angles of a triangle 
in terms of its sides. 



cos" 



But 



\A = 1 — sin 2 \A : 




by (20) 


li i (s — b)(s — c) 

r be 


by (85) 


be — (s 2 — bs — cs -\- be) 




be 




s (b -f- c — s) 






be 




b-\-c = 2s — a, 




b -f- c — s = s — a. 




2 , { s (s — a) y 

cos 2 hA = — ^ J - ; 

be 






cos*iB- s ( s - b \ 
ca 


► 


(86) 


•> i n S (S — C) 

cos- i 0= — ^ —A 

#6 J 







also, 



These formulae may also be derived like those of § 56, by 
starting with the equation, 

V _i_ c 2 — a 2 

COS Yi = ' , 

26c 
and adding both sides to unity, reducing and using (58). The 
student should derive them in this way. 

58. To determine the tangents of the half-angles of a triangle 
in terms of its sides. 

If we divide each one of (85) by the corresponding one of 
(86), we obtain 



76 



PLANE TRIGONOMETRY. 



tan' jA = <*-*> («-">,] 

s (s — a) 

tan' 13 =.<«-«) ('-«>, 

« (* — 6) ' 

tmiMC= ( *- a)( *-* ) . 
s (« • — c) J 



(87) 



These formulae may be reduced to a more convenient form 
by the following substitution : 



Let 



4 



(s — a) (g — 6) (g — c) 



From (87), tan 2 \A = (s - a) (s - b) (s- c) 
y J s (s — ay 



or 



also, 



tan 2 \A 
tan \A 
tan \B 
tan \C 



(s — a) 2 

r 
s — a ' 

r 
s — &' 



59. Area of a Triangle. 

Let iT represent the area of 
any triangle, ABC, Fig. 22. 
We know from geometry that 

but, 

p = b sin A — a sin B : J. 

K= Ibc sin A = \ac sin i?, 
from which, by analogy, 

K = \ab sin C. 



Fig. 22. 



(88) 



(89) 




(90) 



*See I 68, equation (93). 



SOLUTION OF OBLIQUE TRIANGLES. 



77 



Another value for K can be derived as follows : 
Since by (54) sin A = 2 sin I A cos I A, 

we have, from the first of (90), 

K = bc sin $A cos \A ; 



but s 



in ia = \ f {s ~ b ^ s ~ c ^ and cos \A 
\ 6c 



J s(s — a) . 



JT= 1/ s (s — «) (s — b) (s — c). 



(91) 



60. 



EXERCISES. 



Fig. 23. 




1. The formula of $ 54 can be proved geometrically as follows : 

Let ABC be any triangle. Produce 
BC to D, making CD= CA ; and take 
E on ^C so that CE = CA. Then BD 
= a + b, and BE= a — b. AD and AE 
are perpendicular to each other, since 
E, A, and D lie on the circumference 
of a circle whose centre is C, and ED is 
a diameter. Draw EG parallel to AD, 
and hence perpendicular to AE. 

Angle CEA = angle EAC, since the triangle AECis isosceles. 
CEA + EAC=2CEA = 180° — C = A + .B : 

Also angle .EM& = ^4 — EAC= A — CEA = A — %(A 

= HA-B). 
Hence we have 

a + b _ BD = AD _ AE tan AED . 

a — b BE EG AEtsmEAG 1 

a-\-b _ tan %{A + B) 



B) 



a — b tan } (A — B) 

2. The following formulae are sometimes useful : 

a = b cos C + c cos B, 

b = c cos A -{-a cos C, 

c = a cos i? -+- 5 cos A. 
The student can readily prove them by drawing a triangle, and 
letting fall the perpendiculars from the angles to the opposite 
sides. 

3. From the equations in Ex. 2, show that 

a + b + c = (b + c) cos A -f (e + a) cos i? -f (a + b) cos C. 



78 PLANE TKIGONOMETRY. 

4. Prove a + b = ^ os 2 (A — B) = eos h {A — B) ^ 

c sin I C cos J (A + B) 

By (82) «- = - sil ^, and A = ^^ ; 

J v ; c sin C" c sin C 

adding a+A^ sin^ + si n^ 

c sin C 

By (60) and (54) = 2 sin »U + B) cos } (^ - B) 

J K ) K J 2 sin J CcosJC 

_ cos % (A — B) 
sin J<7 ' 
because, since %(A + B) = 90° — JC, sin I {A + 5) = cos JG 
For the same reason sin J-C= cos J (J. + -S)i and we have also 
a ± b _ cos I (A — ^) 
c cos £ (-4 + -#/ 

5. Prove in the same way 

a — b _ sin % (A — B) _ sin j (A — B) 
c cosiC &m${A + B)' 

6. If we add unity to both sides of the equation in Ex. 4, we 

have 

a + b + c _ cos $(A + B) + cos j (A — B) 
c cos £ (A + B) 

By (62) and (16) = 2 cos M cos ^ 

SID 2^ 

Substituting s = J (a + 5 + c), we have 

s cos \A cos \B 

c sin JO 

7. If unity be subtracted from both sides of the equation in 
Ex. 4, we have 

a + b — c = co8 $ (A — B) — cos j (A + B) 
c cos %(A + -B) 

By (63) and (16) = 2 sm ^sin M 

sin 2 k 
or, since a + ^ — c = 2 (s — c), 

a — c sin %A sin JJg 

c sin W 

8. By adding unity to both sides of the equation in Ex. 5, and 
subtracting it, show, as in Exs. 6 and 7, that 

(1) s — b sin \A cos \B ,n\ s — a cos hA sin IB 

U c ~ cos %C ' c — cos JC ~ ' 



SOLUTION OF OBLIQUE TRIANGLES. 79 

9. The four formulae in Exs. 6, 7, 8 may also be derived by sub- 
stituting for the functions ou the right in each their values as 
given in (85) and (86). Derive them in this way. 

10. Show that in any triangle tan A + tan B + tan C= tan .1 
tan B tan C. 

I. We have first (g 42, Ex. 7), 

tan ( 4 ' B-\- C) = * a " ^ ~*~ tan ^ "^ tan ^~ tan ^ tan B tan ^ 
U ^ ; 1 — tany4tan,B — tan.BtanC— tanCtan.4' 

and, since A + .B + C= 180°, tan (^L + 5 + G) = 0. 

Hence tan A + tan i? -f- tan C — tan A tan i? tan (7 = 0, 

or tan .4 + tan B -\- tan C= tan .4 tan B tan C. 

11. Or, as follows. See Ex. 23, p. 57, and (42), 

tan A + tan B + tan (7= sin ^ ~ ^ — tan (A + i?) , 
cos ^4 cos .B v ; ' 

= sin (A + B)[ - 1, 

Lcos.4cos^ cos(v4-f-.B)J' 

= sin ( 1 i £ )[ cos(A + B) — cos .4 cos ^ 1 
Lcos A cos B cos (A + B)i J 

bv f501 = siu M + ^) [ — sin A sin ^ 1 

J V ' cos(^ + ,B)L cos ^1 cos -B J' 

=— tan A tan B tan (A + B), 

by (42) = tan A tan ^ tan C. 

II. Show that if A + ^ + C= 180°, 

tan A J. tan J.B + tan LB tan |C7-f tan AC tan \A = 1. 
12. Show that in any triangle 

sin 2 A -4- sin 2 i? — sin 2 C= 2 sin ^4 sin i? cos C. 
By (42) 

sin 2 ^4 + sin 2 B — sin 2 C= sin 2 A -\- sin 2 i? — sin 2 (.4 — B). 
By (4S) = sin 2 .4 -f sin 2 B — sin 2 A cos 2 B 

— 2 sin A sin B cos A cos Z? — cos 2 A sin 2 i?. 
Now sin 2 ^4 — sin 2 A cos 2 B = sin 2 A (1 — cos 2 B), 

= sin 2 A sin 2 B ; 
and sin 2 B — cos 2 A sin 2 B = sin 2 i? (1 — cos 2 J.), 

= sin 2 A sin 2 i?. 



80 PLANE TRIGONOMETRY. 

Substituting these values we have 

sin 2 A + sin 2 B — sin 2 C 
= 2 sin 2 A sin 2 B — 2 sin A sin B cos A cos B, 
— 2 sin A sin .5 (sin ^4 sin B — cos A cos B), 

by (50) = —2 sin J. sin ^ cos {A + JB) . 

But by (42) —cos (^4 + 5) = cos (7, 

(1) sin 2 A -f sin 2 i? — sin 2 (7=2 sin A sin i? cos (7. 
From this the following are easily deduced : 

(2) sin 2 A + sin 2 B + cos 2 C= 1+2 sin A sin B cos C. 

(3) cos 2 A -f cos 2 i? — cos 2 (7= 1 — 2 sin ^4 sin i? cos C. 

(4) cos 2 ^4 + cos 2 B + sin 2 C= 2 (1 — sin A sin ,B cos (7). 
13. Prove that 

sin A + sin B + sin (7=4 cos M cos JI? cos J (7. 
First, we have by (60) 

sin A + sin B = 2 sin | (^4 + 5) cos | (^4 — jB), 
and by (42) and (54) 

sin C= sin (A + B)=2 sin J (J. + B) cos J (J. + B) ; 

. * . sin A + sin i? + sm C= 

2 sin %{A + B) [cos } (^L + J5) + cos J (^ — -#)]• 
By (16) 2 sin J (^4 + J5) = 2 cos JC, 

and, by (62) cos % (A + B) + cos % {A — B) =2 cos J^4 cos \B, 

sin J. -j- sin B + sin (7= 4 cos JJ. cos Ji? cos J(7. 

EXAMPLES. 
Prove the following relations between the parts of any triangle 



1. c — b cos ^4 = V a 2 — b 2 sin 2 A. 

2. a 2 — b 2 = e {a cos B — b cos ^4). 

3. «^ s i n a=c 



c sin C 

4. cot ^4 + cot B 



ab sin C 



5. cot 4 — cot B 



ab sin (7 



sin (A — B)_ 


_ (a + 6) {a — 6) 


sin (.4 + ^) 


c 2 


a 2 = (6 + c) 2 


sin 2 \A + (6 — c) 2 cos 2 1. 


tan ^4 a 2 — 


- a& cos (7 c 2 + a 2 — 6 2 



tan B b 2 — ab cos C c 2 — a 2 + 6 2 



sill & A 4- B) _ 


_6 + e 


sin %A 


a 


tan (M 4- C) . 


64-c 


tan AJ. 


6 — c' 


tan 1 /I tn.n A 7? 


s — c 



SOLUTION OF OBLIQUE TRIANGLES. 81 

9. 
10. 

11. 

s 

12. 6c cos ^ 4- ca cos 5 4- ab cos (7= A (a 2 4- 6 2 4- c 2 ). 

13. — a cos .1 + 6 cos 2? 4- c cos C = 2a cos B cos C. 

— b cos 2? 4- c cos (7 + « cos A = 2b cos C cos ^4. 

— c cos (7 4- a cos A ~\-b cos 2? = 2c cos A cos 2?. 

14. a cos ^4 4- 6 cos B 4- c cos (7= 2a cos 2? cos C 

4- 26 cos (7 cos A 4- 2c cos .4 cos 2?. 

15. 2«6c sin } (A 4- 5) sin § (A — B) = s (s — c) (a — 6). 

K 2 



16. sin I- A sin £2? sin A (7 = 



abcs 



17. cos^cosAjScos^C:- 



a6c 
18. tan I A tan |5 tan \C= ^-. 



19. cot \A 4- cot £5 4- cot \C 



20. A'= sin \A sin *5 sin AC (— ^— 4- — ^— 4- — ^— V 

\sin-4 sin 2? sin C) 

21. If J. = 2B, then a 2 = 6 (6 + e). 

22. cot \A 4- cot iLB 4- cot W= cot J^L cot %B cot J CI 

23. cot A cot J5 4- cot 5 cot (74- cot (7 cot A = 1. 

24. cos A 4- cos 5 4- cos (7= 1 + 4 sin |^4 sin %B sin J CI 

25. (1) cos 2 A 4- cos 2 B 4- cos 2 (7= 1 — 2 cos J. cos B cos (7. 

(2) sin 2 A 4- sin 2 5 — cos 2 (7= 1 4- 2 cos A cos 5 cos (7. 

(3) sin 2 JL + sin 2 B 4- sin 2 (7= 2 (1 -f- cos ^1 cos B cos (7). 

(4) sin 2 A — cos 2 B — cos 2 (7=2 cos A cos B cos C Y . 

26. (1) sin 2 A + sin 25 4- sin 2(7= 4 sin ^ sin 5 sin C. 

(2) cos 2^4 + cos 25 4- cos 2(7= —1 — 4 cos ^4 cos B cos (7. 

07 a cos ^4 4-6 cos B 4- c cos (7 sin ^4 sin .5 sin (7 

4^ ~ a ."""&"" e .' 

28. 42T= a 2 cot ^4 4- 6 2 cot B 4- c 2 cot C. 



CHAPTER VI. 

SOLUTION OF OBLIQUE TRIANGLES. 

61. The problem of computing the three unknown parts 
(sides or angles) of a plane oblique triangle from the three 
given parts will be divided into four cases, as follows : 

Case I. Given a side and two angles. 

Case II. Given two sides, and the angle opposite to one of 

them. 
Case III. Given two sides and the included angle. 
Case IV. Given the three sides. 

Case I. Given a Side and Two Angles; as A, B, 

and a. 

O) 



First we have 


C=IM° — (A + B). 


Then, by (82), 


b sin B 7 a sin B 
— = -r— -p orb = — — — , 
a sini sm A 


and 


c sin C a sin C 

— = -v—r, orc = — — , 

a smi sm A 



05) 
to 



Note. — There is no check formula for this case that is at 

once convenient and infallible. = == may 

sin A sin B sin G 

be used, but if a mistake is made in finding the logarithm of 
the given side, this formula will not show the error. If the 
required angle be computed from the three sides by the ap- 
propriate one of (87), and the result thus found agrees with 
the result found by (a), the work is correct. 

Example : A = 47° 19 / 18", B = 78° 1^ 42", a = 738.1. 

180° log a = 2.86812 

A + B = 125° 34 7 log sin B = 9.99079 

. • . by (a), C = 54° 26 x colog sin A = 0.13361 

.-.by(/3), b =982.9 log 6 = 2.99252 

82 



SOLUTION OF OBLIQUE TRIANGLES. 



83 



by (y), 



log a — 2.86812 

log sin C == 9.91033 
colog sin A = 0.13361 
log c =2.91206 



c = 816.7 



EXAMPLES. 
Solve the following triangles : 



1. 


B = 70° 16', 


C= 27° 8', 


a = 1.429. 


2. 


^4= 36°3S'28", 


C= 132° 16' 13", 


c = 429.38. 


3. 


A = 58° 19' 12"', 


B= 67° 37' 20", 


c = 38.921. 


4. 


B = 102° 38' 16", 


C= 20° 3' 8". 


b = 47.936. 



Fig. 24. 




62. Case H Given Two Sides and the Angle op- 
posite one of them ; as a, 
b, and A. 

Construct the given angle 
A, Fig. 24, which we suppose 
at first to be less than 90°, 
and lay off on one of its sides 
the distance A C=b. Produce 
the other side indefinitely. 
Now, if a has the value b sin 
A, it will be just long enough to reach from C to the other 
side of the angle A, forming thus a right triangle, ABC. But 
if a < b sin A, no triangle at all can be formed, and hence the 
solution will under such circumstances be impossible. 

If a >> b sin A and < 6, it will be possible to draw a in two 
ways, and hence these will 
be the two solutions, AB X C 
and ABfi. 

If a >> b, there will be but 
one solution, because it is evi- 
dent that a can intersect AB 
only once to the right of A. 

Finally, if .4 > 90°, there 
will be one solution if a > b. and no solution if a <^b. 
glance at Fig. 25 will make this clear. 



Fig. 25. 




84 PLANE TKIGONOMETEY. 

The formulae for the solution of this case are 

t, b sin A . N 

sin B = , (a) 

a w 

C=180 -(A + B), (fi) 

a sin C ( . 

C _ "sin A ' W 

For a check formula for this case, see note on preceding case. 

Ex. 1. A = 38° 54' 42", a = 625.89, 6 = 739.47. 

log b = 2.86892 log a = 2.79649 

log sin A = 9.79804 log sin C x = 9.99933 

colog a = 7.20351 colog sin A = 0.20196 

. •. by (a), log sin B = 9.87047 .• . by (7), log c x = 2.99778 

By (42), there will be two values for B. 

log a = 2.79649 

log sin C 2 = 9.19436 

0.20196 



B x 


= 47° 


54' 


44// 


B 2 


= 132° 


5' 


16" 


A + B, 


= 86° 


49' 


26" 


A + B 2 


= 170° 


59' 


58" 


tc 2 


= 93° 


10' 


34" 


= 9° 


0' 


2" 



• by( ^' |c 2 =9° 0' 2" ' ' \c 2 



.-.by (7), logc 2 =2.19281 

= 994.90 
= 155.89 



Here the given parts, taken with B^ C v and c 1? form one tri- 
angle, and taken with i? 2 , C 2 , and c 2 , they form another. 

Ex. 2. A = 67° 31', a 



= 625.89, 5 = 


= 739.47. 


log b = 


2.86892 


log sin A = 


9.96567 


colog a = 


7.20351 


log sin J? = 


10.03810 



•'•by (a), 

This shows that the solution is impossible, because the number 
whose logarithm is .03810 must be greater than unity, and no 
angle exists whose sine has such a value. (See § 24.) 

Ex. 3. A = 67° 31' 30", a = 625.89, b = 413.14. Here a > 5, so that 
there will be only one solution. This fact also comes out in the 
course of the work, as shown below. 



SOLUTION OF OBLIQUE TRIANGLES. 



85 



log 6 = 2.61610 
log sin A = 9.96570 
colog a = 7.20351 



.*• by (a), log sin B = 9.78531 
.-. JB 1 = 37° 35' 15" 

B, = 142° 24' 45" 

A + B L =105° 6' 45" 

A + B 2 = 209° 56' 15" 

by(j8), C = 74° 53' 15" 



log a = 2.79649 

log sin C = 9.98472 
colog sin J. = 0.03430 
by (y), log c 
c 



i> 2 is impossible. 



== 2.81551 
= 653.90 



EXAMPLES. 
Solve the following triangles : 



1. 


A= 41° 19', 


6=23.61, 


a = 48.72. 


2. 


C = 109° 32' 17", 


c =1026.4, 


6 = 729.5. 


3. 


jfr = 52° 17' 52", 


c = 231.4, 


b = 227.6. 


4. 


A= 58° 37' 36", 


6 = 527.93, 


a = 326.48. 


5. 


C = 72° 36', 


c =47.321, 


a = 12.962. 


6. 


C = 41° 14' 12", 


c =362.41, 


6 = 398.62. 


7. 


^ = 136° 14', 


a = 127.96, 


6 = 134.16. 


S. 


.B = 93° 74', 


6 = 1009.6, 


e = 516.4. 



63. Case ILL Given Two Sides and the included 
Angle ; as a, b, and C. 

Here it is impossible to use (82), for we do not know any 
angle and its opposite side. But (83) can be used, for, while 
neither A nor B is known, their sum is found by subtracting 
C from 180°; and tan J (A — B) is the only unknown quan- 
tity in the formula. 

We have, therefore, for the solution of this ease. 



.4 -f B = 180° — C, 


w 


ni(A — B)~ a — * tan * (J. + tf), 


(« 


A = i(A-\-B) + %(A — Bl 


GO 


B=i(A + B) — h(A — B), 


(*) 


a sin C 

c ~ „,„ A ' 


w 



and for a check, 



6 sin C 



sin i? 



(O 



SG 



PLANE TRIGONOMETRY. 



Example : 


a = 


49.387, b = 41.414 


, C= 109° 38' 36". 




a 




= 49.387 by (a),A + B = 7C 


°21'24" 


b 




== 41.414 






a + b 




= 9O80l 


colog (a + b) =. 


8.04191 


a — b 




= 7.973 


log (a — 6) = 


.90162 


i(A + B) 


= 35° 10' 42" 


log tan i(A + B) = 


9.84810 


by(/2),H^- 


B) 


= 3° 32' 30" 


log tan i(A — B) = 


8.79163 


by (7), A 




= 38° 43' 12" 






by (d), B 




= 31° 38' 12" 


Check. 




log a 




= 1.69361 


log b = 


1.61715 


log sir 


i (7 


= 9.97396 


log sin (7 = 


9.97396 


colog 


sin ^ 


1 = 0.20376 


colog sin B = 
log c = 


0.28023 


by (f) log c 


= 1.87133 


1.87134 


c 




= 743.60 







a = 729.6, 


b = 613.8. 


6=263.94, 


c = 339.62. 


a = 27.92, 


c = 136.91. 


b = 623.12, 


c = 402.09. 


a = 42.61, 


6=43.89. 


a = 622.9, 


c =612.4. 



EXAMPLES. 
Solve the following triangles : 
l.C= 32° 12', 

2. A= 47° 24', 

3. ^ = 127° 38', 

4. ^ = 101° 42' 30' 

5. (7= 92° 31' 24' 

6. B = 59° 31' 12 . 

64. Case IV. Given the Three Sides, a, 6, and c. The 
solution in this case may be performed by the use of (85), 
(86), (87), or (88) and (89). The last set of formula), however, 
are the easiest to apply and give the briefest solution. At the 
same time, they are of equal precision in all parts of the quad- 
rant, while (85) is undesirable if one of the half-angles is 
nearly 90°, and (86) if any are very small (§ 46). We have, 
therefore, 

s = J (a • + b + c), («) 



-4 



(s — a) (s — b) (s — c) 



w 



tan \A 



GO 



SOLUTION OF OBLIQUE TRIANGLES. 



87 





tan £2* = , 
s — 6 


(o) 




tan£C— r , 

5 C 


w 


Check, 


£4 + JjB + £<7=90°. 


(0 


Example: a- 


= 63.89, b = 138.24, c = 121.15. 




a= 63.89 
b = 138.24 
c = 121.15 


colog s = 7.79145 
log (s — a) = 1.99012 
log (s — b)= 1.36922 


Check. 
\A = 13° 45' 35" 
$B = 45° 39 / 0" 


2)323.28 


log (s — c) = 1.60735 
log r 2 = 2.75814 
logr = 1.37907 
log tan \A — 9.38895 
log tan \B = 10.00985 
log tan I C = 9.77172 


£C = 30°35'27" 


s = 161.64 

-s — a= 97.75 
s— 6 = 23.40 
s — c = 40.49 


90° 7 2" 

^4 = 27° 31' 10" 
^ = 91° 18' 0" 
C = 61° 10' 54" 



The student's attention is here called to the fact that with five- 
place logarithms seconds of angles cannot always be accurately 
determined. This accounts for the fact that the sum of the three 
half-angles obtained by the computation above is slightly in 
excess of 90°. 

EXAMPLES. 
Solve the following triangles : 

1. a = 7.296, 6 = 9.821, 

2. a = 121.62, 6 = 16.12, 

3. a = 7249, 6 = 5683, 

4. a = 42.391, 6 = 65.412, 

65. Special Solution for Case H If in Case II. the 
two values of B are nearly 
equal, the method of solution 
there set forth does not give 
these values with very great 
precision. A method of pro- 
cedure for such a case Avill now 
be given. 

Let A, 6, and a be the given 
parts. Let AB X C and AB 2 C 



5.312. 
113.94. 
10002. 
51.833. 



Fig. 26. 




be the two triangles after solution. 



Draw CP perpendicular 



88 



PLANE TRIGONOMETKY. 



to AB 1 and call its length p. 
Let FB X and PB 2 be repre- 
sented by d. We have, first, 

p = b sin A. (a) 

Then, since 

a 2 = p 2 -f- d 2 . 



Fig. 26. 




Taking the value of d with both signs, we then have 



*L*A 



cos B. = — and 


cos i? 2 = , 




(r) 


a 


a 






C 1 = 180° — (A + 5 t ), 


C 2 = 180° — (A + 5 


,)• 


W 


Then, since AP=6 


cos J., 






we have c x = & cos A-\- d and c 2 = b cos JL — <2. 




(•) 


For a check formula, 








a c 


1 _ C 2 




(0 


sin J. sin 


C x sin a,' 


Example : A = 41° 37' 48", 6 = 


13.261, a = 8.832. 








log b 


= 


1.12257 


a = 8.832 


log sin ^4 


= 


9.82237 


p = 8.8092 


.-.by (a), logp 


= 


.94494 


a-j-p =17.6412 


log(a+p) 


= 


1.24653 


a — p = .0228 


log (a —p) 


= 


8.35793 




.'.by (/3), logd 2 


= 


9.60446 


d = .6342 


logd 


= 


9.80223 




log a 


= 


.94606 


B 1 = 85°52 / 56 // 


• '•by (y), log cos B 


= 


8.85617 


^ 2 — 94° 7' 4" 


log 6 


= 


1.12257 


^4 + ^ = 127° 30' 44" 


log cos ^4 


= 


9.87358 


^4 + B 2 = 135° 44 / 52" 


log b cos J. 


= 


.99615 


...by(rf),^ = 52° 29' 16" 
J v ; ' \C 2 = 44° 15' 8" 


6 cos A 


= 


9.9118 


d 


= 


.6342 




..*(.),£ 


= 


10.5460 
9.2776 


CAecZ;. 








log a = .94606 log c a 


= 1.02309 log c 2 


= 


.96743 


log sin A = 9.82237 log sin (\ = 


= 9.89940 log sin C 2 


== 


9.84375 



1.12369 



1.12369 



1.12368 



SOLUTION OF OBLIQUE TRIANGLES. 89 

66. Special Solution for Case EL The following 
solution of Case III. can often be used with more advantage 
than that given in § 03. 

Let the given parts be a, b, and C. Take an auxiliary angle, 

x, so that 

tan x = — , (a) 

b 

from which, by composition and division, 

tan x — 1 a — b . 

tan x -\- I a + b 7 

but, by (73), — X — 1 = tan (x — 45°) : 
' J y y ' tan x + 1 v J 

^=4 = tan (re — 45°). 

a + b K 

If this value be substituted for a , in (/S), § 63, we have 

a+b K J a 

tan | (.1 — B) = tan (re — 45°) tan £ (A -f 5). (0) 

The solution is then completed as before. 

There is a slight saving of labor in this method, because we 
avoid the necessity of finding a + b and a — b and of looking 
up their logarithms. Logarithms of a and b are found at the 
start, and are then ready, when required, at the end for the 
computation of c. If the triangle forms one of a series, and 
logarithms of a and b have been found by previous compu- 
tation, they can be used directly in these formula? without 
finding the actual values of a and b at all. It is desirable in 
the formula for tan x to place the larger of the given sides 
in the numerator, so that tan x shall be greater than 1, and 
hence x greater than 45°. 

Example : a = 76.42, 6 = 51.16, C= 73° 12'. 

log a = 1.88321 

log b = 1.70893 

x = 56° ir 58" .• . by (a) log tan x = 10.17428 

x — 450 = no iv 58 // log tan (3 _ ^ = 9.99666 

i (A + B)= 53° 24' 0" log tan h (A + B) = 10.12921 

I {A — B)= 14° 55' 41" .'.by (3) log tan i (A — B) = 9.425S7 

A = 68° 19' 41" 

B = 38° 28' 19" 



90 PLANE TRIGONOMETRY. 

Check. 
log a = 1.88321 log 6 =1.70893 

log sin C = 9.98106 log sin C = 9.98106 

colog sin A = 0.03184 colog sin B = 0.20612 



logc = 1.89611 logc =1.89611 

c = 78.72 

EXAMPLES. 
Solve the following examples by the methods of \\ 65 and 



1. 


A = 38° 19' 18", b = 642.13, a = 399.62. 


2. 


B = 59° 28' 50", c = 14.761, 6 = 12.792. 


3. 


(7 = 20° 38' 21", 6 = 6.2896, c = 2.2191. 


4. 


6 = 412.63, c = 311.16, ^L = 98° 17' 12". 


5. 


log a = 1.63891, log 6 = 1.76412, C= 37° 19'. 


6. 


log a = 2.38121, log e == 1.99821, ,B = 103° 18'. 


67. 


EXERCISES. 


1. Given 


i a = 49.337, b = 41.414, (7=109° 38' 6" ; determine K. 


By (90), 


log a = 1.69361 
log b = 1.61715 

log sin O = 9.97396 
colog 2 = 9.69897 




log iT = 2.98369 . • . K = 963.14 



2. Given a = 63.89, 6 = 138.24, c = 121.15 ; determine K. 

By (91), logs =2.20855 

log (s — a) = 1.99012 
log (s — 6) = 1.36922 
log (s — c) = 1.60735 
log iT 2 = 7.17524 
log K = 3.58762 . • . K = 3869.2 

3. Given a — b = 27, A — B = 16°, (7= 74°. Solve the triangle. 
From Ex. 5, p. 78, we have 

c = (a — b) cos|-(7 
sin I {A — .B)' 

^1 + B = 180° — C= 106°. 
By (83) a + E- fr-^.^*) . W 



(a) 



SOLUTION OF OBLIQUE TRIANGLES. 



91 



By (a), 



By 03), 



log {a — 6) = 1.43136 






1.43136 


log cos £C = 9.90235 


log tan £ 


{A + B) = 


10.12289 


colog sin \ (A — B)= 0.85644 


colog tan 


i(A-B) = 


10.85220 


log c = 2.19015 


log (a + 6) 


2.40645 


c = 154.94 




a + b = 


254.95 






a — b — 


27.00 


A + B = 106° 




.-.2a = 


281.95 


A — B= 16° 




26 = 


: 227.95 


2.4 =122°, A = 61 c 


> 


a = 


- 140.97 


2,8 = 90°, i? = 45< 


j 


6 = 


113.97 


4. Given K= 17836, ^ = 58° 12', 


6 + e=^638.2. 




From (90) be — . 2jK ' . 

sin JL 




(a) 


(6 — c) 2 = (6 + c) 2 — 46c 




w 


By (a), log 2 = 0.30103 


By 09), 


log (6 + c) 


= 2.80496 


log A" = 4.25129 




log (6 + c) 2 


= 5.60992 


colog sin A = 0.07064 




(b + c) 2 


= 407310 


log be = 4.62296 




46c 


= 167890 


log 4 = .60206 




(6-c) 2 


= 239420 


log 46c = 5.22502 








6 + c = 638.2 




log (6 — c) 2 


= 5.37916 


6 — c = 489.3 




log (6 — c) 


= 2.68958 


26 = 1127.5, 6 = 


. 563.75 






2c = 148.9, c = 


74.45 







Now 6, c, and A are known, and the triangle can be solved by 
Case III. 

5. Given one angle and the opposite side, with the perpendicu- 
lar from the given angle to the given side, to solve the triangle. 

Let the given parts be C, c, and h. Draw a triangle and let fall 
the perpendicular CD = h. Then we have 



sin A 



and sin B = — 
a 



From (82) sin A 



sin 


A 


sin 


B = 


ab' 








_a sin 

c 


C 


an 


d sin 


B: 


_6 


sin 

c 


C 


sin A 


sin 


B 


__ab 


sin 2 


C 







92 PLANE TRIGONOMETRY. 

Equating these two values of sin A sin B, we have 

K 1 ab sin 2 C . 

ab ~ c 2 

ab = ^-. (a) 

sin C 

From (84) c 2 = a 2 + 6 2 — 2a& cos (7 ; 

a 2 + b 2 = c 2 + 2a& cos <7. (0) 

Having determined ab and a 2 + b 2 by (a) and (/3), we readily find 
a -f b and a — 6 ; and then a and 6. The rest of the solution 
presents no difficulties. 

EXAMPLES. 

1. Find the areas of all the triangles given in the examples at 
the end of \\ 63 and 64. 

2. The diagonal of a j)arallelograrn is 500, and the angles it 
makes at one end with the sides are 46° 36' and 10° 12'. Find the 
sides of the parallelogram, the other diagonal, and the angles 
which it makes with the sides. 

3. In an isosceles trapezoid the length of a diagonal is 237, and 
at one end it makes with the base an angle of 18° 12', and with the 
side an angle of 37° 27'. Find the lengths of the four sides. 

4. The parallel sides of a trapezoid are 628 and 417. The angles 
at one end are 62° 13 / and 117° 47', and at the other 84° 38' and 95° 
22 / , the non-parallel sides forming in each case acute angles with 
the longer one of the parallel sides. Find the lengths of the non- 
parallel sides. 

5. A line through the vertex of an equilateral triangle divides 
the angle into two parts, which are in the ratio 2 : 3. Find the 
ratio in which it divides the opposite sides. 

6. One side of a parallelogram is 29, and the angles which the 
diagonals make with that side are 36° and 54°. Find the length 
of the other side. 

7. The diagonals of a parallelogram are d = 102.12 and d x = 
141.16, and their included angle <p = 72° 12'. Find the sides and 
angles. 

8. The area of a parallelogram is 26,112 square feet, and the 
diagonals are 204 and 298 feet. What are the sides and angles? 

9. Find the distance between the objects mentioned in Example 
62, p. 71, if the lines joining them with the foot of the tower make 
an angle of 78° 12 r with each other. 



SOLUTION OF OBLIQUE TRIANGLES. 93 

10. To find the distance between two inaccessible points, il/and 
N, a surveyor measures a line, AB, 350 feet long. At A he finds 
the angle BAM= 102° 19' 18" and the angle BAN = 41° V 12", and 
at B the angle ABN= 98° 16' 24" and the angle ABAT= 52° 17' 48". 
Determine the length of MN. 

11. The lines AC and i?C, joining the foot of a tower, C, with 
two points, A and B, in the same horizontal plane with C, make 
with AB the angles ABC= 78° 12' and BAC= 93° 16'. The angle 
of elevation of the tower at B is 17°. If AB = 127.2 feet, find the 
distance J.C, the height of the tower, and its angle of elevation 
at A. 

12. When the elevation of the sun is 48°, a pole, standing verti- 
cally on a slope whose angle with the horizon is 15°, casts a shadow 
directly down the slope 44.3 feet long. How high is the pole? 

13. A pole stands upon a slope of 7°. From two points on the 
same side of the pole, 150 feet apart and in the same vertical plane 
with the pole, the angles of elevation of its top are 23° 9' and 35° 
12' respectively. How far is the pole from the nearer point and 
what is the elevation of its top above that point? The given and 
the required distance are both supposed to be measured horizon- 
tally. 

14. Give the general solution of the above problem if the angles 
of elevation are § and 6 (^ being the less), a the slope of the ground, 
and d the distance between the points. 

15. Two circles whose radii are 31 and 23 intersect. The angle 
between their tangents at the point of intersection is 27° 30'. Find 
the length of the line joining their centres. 

16. Three circles whose radii are 12, 16, and 25 are tangent ex- 
ternally. Find the angles between the lines joining their centres. 

17. In a triangle A = 46°, 6 = 121, c = 157. The side a is tri- 
sected and a line is drawn from A to the point of trisection of a 
that is nearer B. Find the length of this line. 

18. To find the distance between two inaccessible points, A and 
B, a line CD is measured so that CD cuts AB. The length of CD 
is 600 feet. The angle DCB = 58° 12', CDB = 49° 38', ACD = r i±° 
16', ADC= 62° 13'. Find AB. 

19. To find the width of a river a point B is located on one 
bank and a line AC, 550 feet long, is measured on the other bank, 
parallel to the stream. If the angle CAB = 56° 38', and ACB = 
65° 2', what is the perpendicular distance from B to AC? 

20. To find the distance between two inaccessible points, A and 



94 PLANE TRIGONOMETRY. 

B, a point C is taken in the line joining A and B, and a point D 
without this line. If CD is 437 feet, angle DCB = 114° 42', ADC 
= 65° 7', and CDB = 41° 21', what is the length of ^4jB? 

21. Two points, A and i?, are on opposite sides of a stream, and 
on account of an island in the stream B is invisible from A. A 
straight line, CD, passing through A is measured along the bank 
of the stream. If CA = 612 feet, AD = 1096 feet, BCA = 71° 12', 
and ADB = 62° 13", determine AB. 

22. Prove that in any triangle K= a% sin B sin C . 

J b 2 sin {B + C) 

Solve the following triangles : 

23. iT= 20602, A = 47°, ^ = 53°. 

24. iT= 20602, a = 214.2, b = 315.8. 

25. JT= 20602, a = 196.2, C=59° 12' 12". 

26. a = 32, b = 25, ^ — B = 23°. 

27. a — b = 42.7, c = 161.2, A — B = 13° 5'. See Ex. 5, p. 78. 

28. a + 6 = 732, J. — ^ = 7° 6', C= 73° 37'. 

29. a + 6 = 527.5, c = 319.3, (7= 58° 31'. 

30. a + 6 = 23.6, A = 48° 19', B = 76° 12'. 

3 1. s = 1219, A = 71° 12', ^ = 26° 32'. See Ex. 6, p. 78. 

32. s = 325.1, — = 1.25, B = 39° 55'. 

33. 8 = 726.9, be = 142600, A = 47° 19'. 

34. a + 6 = 263, c = 92, JT= 4748. 

35. be = 11616, a = 205, J. = 58° 29'. 

36. a 2 — 6 2 = 96.7, A — B = 3° 19', C= 71° 13'. See Exs. 4 and 
5, p. 78. 

37. K= 7298.1, ^4 = 37°, B — C= 13°. See Ex. 22, above. 

38. s = 219.2, K= 7397, A = 65° 16'. 

39. b + c — a = 175.9, ^4 = 48°, (7= 93°. 

40. a-6 + e = 95.2, be = 6398, ^4 = 51° 47'. 

41. a — b = 25, c = 117, .B = 42°. 

42. Given the three altitudes p x = 14.9, p 2 = 1 7.2, p 3 = 21.5. Give 
the formulae for the general solution of this case. 

43. Given a = 41, b =38, and the medial line from Cto AB = 
30. Give the formulae for the general solution of this case. 

44. The three medial lines of a triangle are 13, 19, and 17. Solve 



SOLUTION OF OBLIQUE TRIANGLES. 95 

45. If in Case II. the given angle be 45° and c and c x be the 
values of the required side, show that the angle between the two 
positions of the side opposite the given angle is 

COS" 



e' + c, 2 

46. A triangle has its three sides given, a = 27, 6 = 39, c = 32. 
A line is drawn through B which bisects the area of the triangle ; 
rind the two parts into which B is divided. 

47. In a certain triangle a = 472, B = 58° 19', C=69°31'. A 
line is drawn parallel to a, cutting c and b in A^and F respectively, 
so that BX + CY=XY. Determine XY. Give the formula for 
the general solution. 

48. A line AB is tangent to a circle at A, and a secant through 
B cuts the circle in Cand D. The angle between the secant and 
tangent = 15° 38', BC= 57, BD = 73. Determine the radius of the 
circle and the two chords, AC and AD. 

49. Two points and a straight line are given. The perpendicu- 
lar distances from the given points to the given line are 38 and 51, 
and the line through the points makes with the given line an 
angle of 23°. Find the radius of the circle which passes through 
the given points and is tangent to the given lines. 

50. Three points, A, B, and C, not in the same straight line, 
are inaccessible. A point D is taken on AB produced through B, 
and a point E on AC produced through C. The length of DE 
= 739.2 feet. Angle BDE = 112° 19', CDE = 47° 31', BED = 29° 
26', CED = 57° 25'. Find the distances AB, BC, AC. 

51. Three points, A, B, C, are in a straight line AB = 1000, BC 
= 2000. At a point D in the same plane, the angles ADB and BDC 
are each equal to 35°. Find AD. 

52. If .4 £ = 387 feet, BC= 542 feet, ABC= 
162° 19', APB = 29° 37', and BBC = 51° 31', 
determine the distance of P from each of the 
points A, B, and C. 

[Suggestions : Let BAP = x, BCP = y, and 
T= x + y. Then T= x + y = 360° — o — ft 
— ABC. Then find the value of BP in terms 
of the parts of each of the triangles BAP and 
BCP, equate these two values, and eliminate y by means of T, 
getting 

cota; = cot t[ ^■ ABsind h l"| , y = T— x, etc.] 

LBCsintf> cos T J' u ' J 




96 PLANE TRIGONOMETRY. 

53. A horizontal line is drawn through the base of a tower, 
and three points, A, B, C, are taken in this line. At A the angle 
of elevation of the summit of the tower is a, at B it is 90° — a, and 
at C it is 2a. If AB = a and BC= b, show that the height of the 
tower is 

V{a 2 + b 2 ) — ia 2 . 

54. To find the height of a tower SF, two points, A and B, are 
taken in the horizontal plane through its foot AB = 427.9. The 
angles of elevation of the summit of the tower at A and B are 
respectively 27° 38' and 31° 11', and the angle ASB = 33° 16'. 
What is the height of the tower ? 

55. At each end of a horizontal base of length 2a it is found 
that the angular elevation of a certain peak is a, and at the middle 
of the base it is /?. Prove that the height of the peak above the 
plane of the base is 

a sin a sin (3 

l/sin (/3 + a) sin (/3 — a) 

56. ABCis a triangle, and K is the middle point of AB ; AB 
is drawn perpendicular to BC, cutting CKin L ; show that 

a j- _ ab sin C 
a-\-b cos C' 

57. Two circles, whose centres are at a distance of 76 from each 
other, lie in the same plane but do not intersect. Their external 
common tangent meets the line of centres produced at an angle 
of 15°, and their internal common tangent meets the same line at 
an angle of 37°. Find the radii of the circles. Generalize the 
problem. 

58. Two circles, whose radii are 53 and 31, and the line joining 
whose centres is 72, are tangent to a third circle whose radius is 92. 
Find the angle between the line of centres of the first circles and 
the line through their points of tangency with the third circle. 



CHAPTEE VII. 

MISCELLANEOUS PROBLEMS AND TRIGONOMETRIC EQUATIONS. 

68. The Radius of the Circle inscribed in any 
Triangle. 

Let r = the radius required. Fig. 28. 

We have first 
AE=AF=x, BF=BD = y, 

CB= CE = z. 
Hence 

&(a? + y + *) = a + 6 + c = &, 
or x + y + z = s. 



Then 



x = s — (y + z) = s — a, 

y = S —(z + x)=S — b, 




(92) 



z = s — (x + y) = s — c. 
Now r = ^4.i? tan J J. == .Bi) tan %B, etc., 

r = (s — a) tan J A = (s — b) tan %B, etc. 
If for tan %A we substitute its value taken from (87), we 
have 

r = (s 



v 



«)a/ (5 - 


-b)(s- 


■c) 


a )\ 


s (s — a) 




(s — a) (s - 


-b)(s — 


<0 



(93) 



If the values of tan iA, tan %B, and tan JO be taken from 
(87) and multiplied together, we have 



tan I A tan IB tan iC= — \ 

s \ s 



(5 — a) (3 — 6) (5 — c) 



r = s tan JJ. tan $2? tan J (7. 



(94) 



97 



98 



PLANE TRIGONOMETRY. 



Fig. 29. 



69. The Radius of the Circle circumscribed about 
any Triangle. 

Let B = the radius required. 
From the centre of the circum- 
scribed circle draw OB = B and OE 
bisecting BC in D. This line also 
bisects the arc BEG at E. Hence the 
arc BE measures the angle A. There- 
fore BOE=A. We have then from 
the right triangle BOB 

BD= OB sin BOB, 
or ia = B sin A, 

and similarly ib = B sin B, 

\c = B sin C, 
adding i(a-\-b-\-c) = s = R (sin A -f sin B -f- sin C) 
but, § 60, Exercise 13, 

sin A -j- sin B -j- sin (7=4 cos \A cos \B cos %C, 

s 




B 



4 cos \A cos \B cos \C' 



From (86), we have 
4 cos \A cos \B cos \C . 



(96) 



4s V s (s — a) (s — b) (s — c) 



abc 



B 



abc 



abc 



4 l/ s(s — a) ( S —b)(s — c) 4JT 



(97) 



Example: Find r and B in the example worked out in 
Here r has already been found as an auxiliary quantity by 
which is the same as (93), and log r = 1.37907 : .• r = 23.94. 

By means of (96) we find B as follows : 



64. 



\A = 13° 45' 35" 
^ = 45° 39' 0" 
|C = 30° 35' 27" 

B =69.14 



log s = 2.20855 
colog cos = 0.01265 
colog cos = 0.15550 
colog cos = 0.06509 
coloo: 4 =9.39794 



PROBLEMS AND TRIGONOMETRIC EQUATIONS. 



99 



Fig. 30. 



70. The Radii of the Escribed Circles (or Ex- 
circles). 

A circle drawn as in Fig. 30 
tangent externally to one side 
of a triangle, ABC, and tangent 
internally to the other two sides, 
is called an escribed circle, or 
excircle. Three such circles can 
be drawn with any triangle. Let 
the radius of the one in the 
figure be r v The centre of this 

circle lies on the bisector of A, and also on the bisectors of 
the exterior angles at B and C. 




CQ) 



AP=AB -j- BP=AB -f BQ, 
AB = AC + CB = AO + CQ, 
AP + AB = AB -f A C + (BQ 
= a + b + c = 2s. 
But AP=AB, .-. AP = s. 

Hence, from the right triangle AOP, 

r t = s tan iA. (98) 

By analogy we have for the radii of the other two escribed 
circles, 
r — s tan IB and r„ = s tan JC 



71. To determine a point X 
within a triangle, such that the 
angles XAB, XBG, and XCA shall 
be equal, and to show that, if d be 
this angle, 

cot 8 = cot A -f cot B -J- cot C. 
From the figure : 

AXB = 180° — d — (B — d) 

BXC = 180° — C, 

CXA = 180° — A. 




180° — 5, 



100 



PLANE TRIGONOMETRY. 



From the triangles AXB and 
AXC: 

, y_ csin(2?— -3) _ csin(ff— d) 
~~ sin ^4 JT£ ~~ 



and AX= 



sm 



sin i? 

6sin<5 



sin yl JTO sin A ' 
c sin (B — d) b sin 8 



or 



But 



sin B 

sin (5 



sin A' 




sin _B 



sin B sin <5 c sin A sin J_ sin C 
sin (5 — 5) sin B cos 5 — cos B sin 



sin B sin 



sin i? sin 



= cot 



by (82) 



cot B. 



sin B 



sin 01 + 0) = CQtA Qt c 
sm A sin sm A sin C 

cot <5 — cot B = cot J. -j- cot C, 
or cot d — cot A -\- cot B -\- cot 0. 

The point thus determined, and the corresponding point 
determined by taking XA C = XBA = XCB = d, are called 
Brocard's points ; and around them centre some of the most 
interesting developments of modern geometry. 

72. To solve the equations 

m sin cp = 
m cos <p 

where a and b are known and the values of m and <p are re- 
quired. 

Dividing the first equation by the second, 

, a 

tan (f = — 
b 

.-. when <p is known, 






(99) 



(«) 



m = 



— or m = . (/S) 

sm (p cos (p 

This solution, it will be observed, is always possible what- 
ever values a and b may have. As determined by (a), <p will 
have two values, differing by 180°. If a and b have the same 



PROBLEMS AND TRIGONOMETRIC EQUATIONS. 101 

sign these values will be in the first and third quadrants ; but 
if a and b have unlike signs they will be in the second and 
fourth quadrants. In either case, therefore, sin <p and cos <p 
will each have two values numerically equal, but with opposite 
signs. Therefore m will have two values numerically equal, 
with opposite signs. 

If certain restrictions be laid upon either m or <p the equa- 
tions give only single values for the unknown quantities. 

1st. Let m be positive. Then sin <p and cos <p must have the 
signs of a and b respectively, and hence the quadrant of <p is 
determined. 

2d. Let (p be < 180°. Then the quadrant of <p is deter- 
mined, and hence the signs of sin <p and cos <p are determined. 
Therefore m will have only one value, which will be -f- or — 
according as the signs of a and b are like or unlike those of sin 
<p and cos <p respectively. 

3d. If <p is limited to values < 90°, either 4- or — , the signs 
of sin <p and cos <p are determined, and hence the value of m, 
as before. 

Example: Given m sin = - 
and <p {<p to be < 180°). 
We have, by (a) and (/3), 



<j> = 117° 52' 19' 



■72.631, m cos $ = 38.412. Fir 


id m 


log m sin <j> = (— ) 1.86113* 




log m cos = 1. 58-146 




log tan ^ = (— ) 10.27667 




log m sin <p = (— ) 1.86113 




log sin (p = 9.94645 




logra =(— ) 1.91468 





m=— 82.164 

73. To solve the equation 

a sin x -f b cos x = c, (100) 

where <2, 6, and c are known and x is required. 
Assume m sin y> = 6, ) 

m cos <p = a, ) 

* The minus sign is written thus ( — ) to remind us that this is the loga- 
rithm of a negative quantity. It of course does not affect the logarithm 
itself in any way. 



(«) 



102 PLANE TRIGONOMETRY. 

tan <p = — , m = , or m = . (p) 

a sin <p cos <p 

Substituting (a) in (100), we have 

m sin x cos <p -j- m cos x sin <p = c, 

or, by (48), m sin (x -{• <p) = c, 

or sin (# + <p) = — (101) 

m 

by which # is found when m and ^ are known. 

If in (101) we substitute the values of m found from (/5), 
we have 

, , N c sin <p c cos <p , 1nm 

sin (x + <p) = — j-z- = — — x., (102) 

by the use of which the necessity of finding m is avoided. 

Note. — If (101) be used, it will be convenient in solving (a) 
for m and <p to restrict m to positive values; but if (102) be 
used, (p may be restricted to values <[ 180° without regard to m. 

Example : Given — 23.8 sin x -f- 19.3 cos x = 17.5, to find x. 
a = —23.8, 6 = 19.3, e = 17.5. 









log b 




= 


1.28556 








log a 




= (- 


-) 1.37658 


+= 


140° 57' 38" 




log tan <j> 

logc 

log sin <p 




= (- 


-) 9.90898 

1.24304 

9.79924 


fh — . 


f 34° 49' 40" 
{ 145° 10' 20" 




colog b 
log sin (a: 


+ 0) = 


8.71444 


r 


9.75672 




x = —106° 


7' 


58" and 4° 


12' 


42", 






x= 253° 


52' 


2" and 4° 


12' 


42". 





or 

74. To solve the equation 

sin (<p -\- x) = a sin x, (103) 

in which <p and a are known, and x is required. 

From (103) we have 



sin (<p 4- x) 
sin x 

sin (<p -f- .t) -f- sin # a-f 1 

sin (?? -j- a') — sin # a — 1 



(«) 



PROBLEMS AND TRIGONOMETRIC EQUATIONS. 103 

By (64), Bin (<p -f x) -f sin or = tan (x + ?? Q 

sin (^ -f- #) — sm # tan 2^ 

and if we let tan y = a, 

we have £±J = tan r + * = cot ( r — 45°). [by (73)] 
a — 1 tan y — 1 

Hence (a) becomes 

tag (x + jy) = cot( 45Q), 



(* + irt = cot (r— 45 °) tan *?>! , 104) 

tan y = a, J 



tan i^> 
or tan 

which, with 
fHves the solution. 

Example : Find a; from the equation 

sin (106° + a?) = —1.263 sin x. 
a = —1.263, log a = (— ) .10140 = log tan y. 
y = 128° 22' 16" 

y — 45° = S3 22' 16" log cot (y — 45°) = 9.06526 

!0 = 53° 0' 0" log tan ^ = 10.12289 

a: 4- '* = 1 8 ° 46 ' 2 " l0g tan ( * + W = 9 - 18815 

" rw \l88°46' 2" 

a; = —44° 13' 58" and 135° 46' 2", 
or x= 315° 46' 2" and 135° 46' 2". 

75. Tb so?ve Me equation 

sin (^ — a:) = a sin x, (105) 

where ^ and a are known, and x is required. 

Let the student show that the solution of this equation is 
found by 

(106) 



tan y = a, 



tan (# — Jp) = cot ( r -f 45°) tan h, 

76. To solve the equation 

tan (^ -f- x) = <2 tan a?, (107) 

where ^ and a are known, and £ is required. 

From (10 - — v ' ' J - = a, 

v J tana; 



and, hence 



tan (<p -\- x) -{- tan # a -f- 1 

tan (c? -j- x) — tan x a — 1 



104 PLANE TRIGONOMETRY. 

By (70), 



tan (<p -\- x) -j- tan x 


sin ($0 -j- 2x) m 


tan (<p -{- x) — tanx 

sin (<p -j- 2x) _ 


sin y> 
_a-\-l 


sin <p 

74, 


a — 1 



(108) 



or, as in 

sin (<p -j- 2x) = cot (j — 45°) sin <p, 
where tan y = a. 

Example : Find x from the equation 

tan (23° 16' + x) = .296 tan x. 
a = .296, log a = 9.47129 = log tan y. 

7 = 16° 29' 20" 

7 _ 45° = —28° 30' 40" log cot (y — 45°) = (— ) 10.26503 

(p = 23° 16' 0" log sin ^ 9.59661 

<> 4- 2a; = i 226 ° 39 ' °" log sin ^ + 2x ^ = (~~) 9.86164 

9 ~ t 1313° 21' 0" 

2^ = 203 o 23 r 0" and 290° 5' 0"; 
and a; = 101° 41' 30" and 145° 2' 30". 

77. 2tf so/ve £/ie equation 

tan (^ + #) tan x==a, (109) 

where ^ and a are known, and # is required. 
From (109) we readily deduce 

1 -j- tan ((p -f- x) tan x 1 -j- a 

1 — tan (<p -J- #) tan sc 1 — # 

-o ,-,... 1 4- tan (<p j- a?) tan # cos y> 

1 — tan Qcp -f- #) tan x cos (^ -f - 2a;) ' 

°°L2 = izM = tan ( + 450) b (72) 

cos (p + 2x) I — a Kf ^ h J K J 

or cos (jcp -j- 2x) = cot (j -j- 45°) cos <p,' 

where tan y — a. 



(110) 



78. ^o sofee #ie equation 

a = a -[■• b sm a - 
Such an equation in which a and b are known, and a is 
unknown, can be solved only by a process of successive 
approximation. The method is exemplified below. 



PROBLEMS AND TRIGONOMETRIC EQUATIONS. 105 

Let a = 63° 19', b = 38° 17'. Then the equation is a = 63° 
19' -f 38° 17' sin a. Since the value of sin a is between 1 and 
— 1, a must lie between 63° 19' — 38° 17' = 25° 2' and 63° 19' 
-j- 38° 17' = 101° 36'. Hence sin a is positive, so that a is 
further limited to values between 63° 19' and 101° 36'. Since, 
therefore, sin a will be nearly 1, a will be somewhere near 90°. 
Let us try 

a' =85° .-. log sin a' = 9.99834 

b = 38° 17' = 2297' .-.log b =3.36116 

b sin a' = 2288'.2 log b sin a' = 3.35950 
a = 3799'.0 



a -f b sin a' = 6087'.2 = 101° 27'.2 

The assumed value of a (85°) is, therefore, too small. More- 
over, since sin 95° = sin 85°, 95° would be also too small ; and, 
since if we make a larger than 95°, sin a becomes smaller, the 
true value of a must lie between 95° and 101° 27'.2. 

Let 



a" 

b sin a" 


= 98° 
= 2274'.6 
= 3799'.0 


101° 


.-. log sin a" =9.99575 
log b =3.36116 


a 


log b sin a" = 3.35691 


a -J- b sin a' 


= 6073'.6 = 


13'.6 



It will be observed that an increase of 3° in a diminishes 
the value of a -\- b sin a by only 13'.6 ; we can, therefore, now 
guess very nearly the correct value of a. 

Let a!" =101° . • . log sin a'" = 9.99195 

b sin a'" = 2254'.8 loar b = 3.36116 



a = 3799'.0 log b sin a'" = 3.35311 

a + b sin a"' = 6053'.8 = 100° 53'.8 

This approximation is, therefore, too large. 

Let a- = 100° 54' • . log sin oT = 9.99209 

b sin a- = 2255'.5 log b = 3.36116 

a = 3799'.0 log b sin oT = 3.35325 

a + b sin oT = 6054'.5 = 100° 54'.5 



106 PLANE TRIGONOMETRY. 

Hence a = 100° 54'.5 log sin a = 9.99208 

6 sin a = 2255'.5 log 6 = 3.36116 

= 3799'.0 log b sin a = 3^5324 



a 



a + bsma = 6054'.5 = 100° 54'.5 
which verifies the result. 

EXAMPLES. 

1. Find r in each of the four triangles given at the end of § 64. 

2. Find R in each of the same triangles. 

Find the values of the literal quantities in the equations that 
follow : 

3. a sin x = 4.296, a cos x = — 4.782, where a is to be positive. 

4. p sin = .036289, p cos $ = .013462, where </> is < 180°. 

5. m sin = —42.93, m cos = — 108.16, where m is positive. 

6. r sin = —4.261, r cos ^ = 8.316, where 6 is < 180°. 

7. 16 sin a; — 42 cos x = —12.92. 

8. .0621 sin x + .6831 cos x = .6738. 

9. 21.14 sin (x + 15° 120 - 16-42 cos (a; + 15° 12') = 20.12. 

10. 38.96 sin (x — 108°) + 14.62 cos {x — 108°) = —25.98. 

11. 21 sin (x + y) — 38 cos {x + y) = 40, 
16 sin (x — y) + 9 cos (x — y) = 12. 

12. Show that the solution of 

a sin a? + 6 cos x = c 
is impossible if c 2 > a 2 + 6 2 . 

Find a; from the equations that follow : 

13. sin (26° — x) = 18 sin x. 

14. sin (138° 16' + x) = —.04623 sin x. 

15. tan (93° 18' + *) = —2.689 tan x. 

16. tan (58° + x) tan x = 1.6821. 

17. sin (37° 19' + x) = 4.263 sin x. 

18. tan (46° 12' + x) = —.6381 tan x. 

Derive general formulae for solving the following equations, as 
in U 74-77, m and <p being known in each case. 

19. cos {<p + x) = m cos x. 20. cos (<j> — x) = m cos x. 
2 1. tan — x) = m tan #. 22. tan (0 — x) tan a; = ra. 
Apply the results just found to solving : 

23. cos (38° 19' + x) = .63 cos x. 



PROBLEMS AND TRIGONOMETRIC EQUATIONS. 107 

24. cos (17° — x)= —.428 cos x. 

25. tan (63° 17' — a;) = 14.728 tan x. 

26. tan (138° — x) tan x = —1.4263. 

27. Find ?* and <p (r to be positive) from r sin (# + 12°) = 18.14, 
rcosty + 12°) =—16.12. 

28. Find m, n, x, and y {m and n to be positive) from 

m sin (x +y)= 46.12, w sin (x— y) = —13.41, 

m cos (a? + #) = —76.12, n cos (a; — y) = 26.91. 

29. Find m and a; from 

m (sin a; + cos a:) = —128.12, 
m (sin x — cos a*) = 14.64. 

30. Find r, <}>, and from the following equations, with the con- 
ditions that r shall be positive, and < 180°. 

r cos sin = 2.683, 
r sin (j> sin = — 4.123, 
rcos0 = 6.211. 

31. Also, with the same conditions, from 

r cos <p sin = —6.896, 
r sin sin = .412, 
r cos = 4.216. 

32. If p x , p 2 , and p 9 be the three perpendiculars let fall from 
the vertices of a triangle to the opposite sides, show that 

Pi 1H Ih r 
where r is the radius of the inscribed circle. 

33. Show that 

{a + & + C ) (J- + JL + — ) = (Pi + P* + Ps) ( — + — + — ) • 
\ a b c I \ Pi P% Pit 

34. Show that p x p 2 p z = -^- sin 3 (7. 

35. Show that s = r (cot *^4 + cot IB + cot W). 



36. Show that i? = i-v/ , abc „ . . 

2 * sin .4 sin i? sin C 

37. Show that a& = 2B 2 [cos (^4 — B) — cos (J. + By]. 

38. Show that a 2 + b 2 = 2B 2 (2 — cos 2.4 — cos 2B). 

39. Show that a 2 + b 2 = ±B 2 [1 + cos C cos (.4 — jB)]. 

40. Show that 

a j? & + c 5 — c 

sin h(X+C) cos H^B— G) cos £ (# + C) sin § {B — C)' 



108 PLANE TRIGONOMETRY. 



41. Show that R = x/r" . ^ etc. 

4 cos \A sin \B sin J (7 

42. Show that r = AR sin \A sin Ji? sin JC. 



43. Show that /-!= y^ — ^ — -*, r 2 = etc., r 3 = etc. 

44. Show that i\ = ±R sin \A cos ii? cos £C, etc., etc. 

45. Show that r x + r 2 -f r 3 = 4i? — r. 

46. Show that -L + J_ + J_ = J-. 

1\ r 2 r 3 r 

47. Show that rr x r 2 r z = K 2 . 

Solve the following triangles from the parts given. 

48. R = 104.96, a — 36.21, b = 71.39. 

49. i? = 91.28, A = 23° 36', ^ = 78° 12'. 

50. r = 14.69, ^ = 49° 38', <7= 100° 12'. 

51. r = 62.93, (7=121° 13', a = 193.8. 

52. 6 + c = 605.9, R = 319.04, A = 31° 4' 48". [Use Ex. 40.] 

53. b + c = 6.928, J? = 2.5, a = 4.762. 

54. s = 131.25, i? = 53.82, ^ = 65°. 

55. b — c = 1.293, i? = 13.726, A = 53° 6'. 
50. a + b — c = 62.14, i? = 50.71, A = 101° 9'. 

57. a + 5 — c = 72.3, i? = 61.17, C= 75° 38'. 

58. K= 52963, i? = 250, a = 326. 

59. s = 88.92, r = 9.73, A = 110° 19'. 

60. b + c = 1060, r = 123.79, a = 349. 

61. a + 6 = 389.2, r = 49.6, C= 70°. 

62. r = 25, a = 326, A = 129° 16'. 

63. r = 10, R = 29, a = 40. 

64. r = 3.92, i? = 10.18, A = 83° 35'. 

65. R = 62, a 2 — 6 2 = 4000, J. = 80°. 

66. i? = 13, a + 6 = 29.2, ^L — 5 = 17°. 

67. R = 9.36, a& = 152.93, A = 58° 3'. 

68. r = 10.5, s = 59, a = 41. 

69. x — 20° cos x = 32°. Find x to the nearest minute. 

70. 6 = 37° 58' + 3° 27' sin 0. Find to the nearest minute. 

71. a -f 5° 3' sin a = 83°. Find a to the nearest minute. 

72. Show that in any triangle the sum of the segments of the 
three altitudes, formed by their point of intersection, which are 
adjacent to the angles, is equal to 2 (R + r). 



CHAPTEE VIII. 

DE MOIVRE'S THEOREM, TRIGONOMETRIC SERIES, HYPERBOLIC 
FUNCTIONS. 

79. In this chapter the imaginary unit, -j/ — 1, will be used 
extensively. In accordance with the general custom this 
quantity will be represented by i. As the student has already 
learned in algebra, the integral powers of i form a cycle of 
four recurring values. Thus 

i=|/ — 1, i 2 = — 1, i 3 = — -[/ — 1 = — i, i 4 =lj ?=-]/ — 1 = '', etc. 
The following theorem is also proved in algebra : 

If x -f- yi = a -f- hi, then x = a and y = b. 
Stated in words, this theorem says : If two expressions, each 
composed partly of real and partly of imaginary quantities, are 
equal, then the real parts of the two expressions are equal to each 
other, and the imaginary parts are equal to each other. 

80. De Moivre's Theorem. If the two expressions 

cos 6 1 -j- i sin X and cos 2 -f- i sin 2 
be multiplied together, the result is 

(cos X -f i sin X ) (c.os 2 -f- i sin 2 ) = cos X cos 2 — sin X sin 6 2 
-f- i (sin X cos d 2 -f- cos X sin 2 ), 

= cos (0 X + 2 ) -J- t sin (0 X + 2 ). 
Similarly 

[cos (0 X -f 2 ) -f- i sin (^ -f- 2 )] [cos 3 -f- i sin 3 ] = 

cos (^ + 2 + 0,) + i sin (O x + 2 + 3 ). 
And, in general, the product of n factors of the form 

cos -\- i sin 
is cos (*! + *,+ ..... O + i sin (^ + 2 4- . . . . n ). 

The following corollaries are readily deduced from the 
general theorem : 

10 109 



110 PLANE TRIGONOMETRY. 

I. Since cos — i sin 6 = cos ( — 0) -\- i sin ( — 0), we have 
(cos X -j- i sin X ) (cos 2 — i sin 2 ) = cos (0 X — 2 ) -j- i sin (d l —d 2 ). 
If now we make X = V this equation reduces to 

(cos 0j -|- * s i Q ^i) (cos 2 — i sin 0J = 1. 

II. Since 

1 _ cos — % sin 

cos -\-i sin (cos -\- i sin 0) (cos — i sin 0)' 
by I. == cos — i sin 0, 

we have 0OS e i + im n Ji = ( CO s 0. + j sin 0.) (cos 2 — i sin 2 ) 
cos0 2 +z*sin0 2 v * . 2 y 

= cos (0, — 2 ) -f i sin (0 : — 2 ). 

III. If in the general equation for the product of n factors 
of the form cos -j- i sin we make all the factors equal to 
each other, we have 

(cos -J- i sin 0) n = cos n0 -j- i sin n0, (HI) 

where n is supposed to be a positive integer. This equation 

reduces to 

i 

cos -j- i sin = (cos n0 -(- i sin n0) n . 

Hence, if we write in place of ~n0, and therefore — for 0, 

n 

we have 

- 

(cos 4- i sin 0) n :=cos- f- i sin — . 

n n 

If we raise both sides of the last equation to the with 
power, where m is a positive integer, we have 

— / \ m m m 

(cos 0-\-iam0) n -=[ cos 1- i sin — ) = cos — A- i sin — 0. 

\ n n ) n n 

Hence (111) is true if n is a positive rational fraction. 

To extend the proof to the case of negative exponents, we 
have 

(cos 4- i sin 0)-" = : = (cos — i sin 0) n by II. 

v ^ J (cos + % sin 0)* J 

= [cos (— 0) -f i sin (— 0)~\ n = cos ( — ?i0) -f i sin ( — n0) 

= cos n0 — i sin n0. 



DE MOIVRE'S THEOREM, ETC. Ill 

It appears, therefore, that (111) holds true for all rational 
values of n. 

81. We have written above 

A ft 

(cos -j- i sin 0)n = cos 1- i sin — . 

n n 

Now, on the left, 6 may be replaced by 2A'tt -j- 0, where k is an 
integer, without altering the value of the expression. Hence 

we have 

i_ i_ 

(cos d -f- i sin 0)» = [cos (2A'tt -f 0) -f i sin (2A- -\- 0)]« 

2for + . . . 2A- + 6> . 110N 

= COS — \- I 8111 ^^ (112) 

?l ft 

If we give to k each of the values 0, 1, 2, ... . (n — 1), in 
turn, this last expression will have a different value for each 
value of A, and hence n values in all. It cannot have more 
values than w, because if k = n -j- p, where p <^n and integral, 
we have 



2 (n + p) - + , . . 2 (n 4- p) r + 

cos — * — ] -^- 1 ! (- i sm — * — L^v 1 — 

n n 

= C08 ( 2 , + 2 P* + ") + i sin (2, + »» + *], 

= C0S ( J hr-) + ism ( J h^)' 

which is the same as for k =p. 

From this it follows that every expression cos d -{- i sin 
has ft nth roots. This is equivalent to saying that every 
quantity has n nth roots, as is shown below. 

(1) Every quantity can be expressed in the form x -\- yi. 
This will represent a real quantity if y = 0, a pure imaginary 
if x = 0, and a complex quantity if neither x nor y is 0. 



(2) 



yi = Vx* 4- fi x 4- 1 y \ . 

\V x 2 4- y 2 Vx 2 4- y 2 ) 



If we put — " = cos and — ^ = sin <?, which we 

l/^r 2 + y 2 Vx 2 + ?/ 2 



112 PLANE TRIGONOMETRY. 

may do, since the sum of the squares of these quantities is 
equal to unity, we have 

x -j- yi = Vx 2 -j- y 2 (cos -f i sin 0), 

where 6 = tan -1 ^-. 

x 

(3) By (112), taking the nth root of x -f- z/a, we have 

l^+Ji = V*+? (cos 2k * + 6 + i Bin *±±±\ 
\ n n ) 



2m 



Hence if we multiply the numerical value of vx 2 -f- z/ 2 by 
each of the n values of the second factor, we shall have the 



n values of y x -f- yi. 

In an expression like the foregoing x -\- yi = V x 2 -f- y 2 
(cos 6 -\- i sin 0). the positive value of V x 2 -j- y 2 is called the 
modulus of the complex quantity x -j- yi ' ; and 6 is called the 
amplitude or argument of the same quantity. In the case of a 
real quantity the modulus is always equal to the quantit}^ 
itself with the positive sign ; while the amplitude is if the 
quantity is positive, and n if it is negative. The modulus of 
a pure imaginary is the coefficient of i with the positive sign ; 

and the amplitude is — if the coefficient of i is positive, and 

Q 

— if the same coefficient is negative. 

EXERCISES. 

1. Find the three cube roots of — 8. 

—8 = 8 (—1 + i . 0) = 8 (cos tt -f i sin tt), 

if k = =2 (cos 60° + *sin 60°)= 2(J + | 1 /3.*j 

if k == 1 =2 (cos 180° + i sin 180°) = —2 

if & = 2 =2 (cos 300° + « sin 300°) = 2 (| — § t/3 . ?'). 

2. Find the square roots of 2 -+- 2 /3 . i. 

2 + 2 1 /3.« = 4(£ + £ 1 /3.*) = 4 (cos-|- + *sin-|-), 



DE MOIVKE'S THEOREM, ETC. 113 

, / 2&7T + — 2&tt + -^\ 

v2 + 2i/3.i=i/4 I cos_ 3_+2Sin_ 3l 

V 2 2 / 

if k = =2 (cos 30° + i sin 30°) = ^3 + * 

if A = 1 =2 (cos 210° + z sin 210°) = — t/3 — i. 

3. Show that the three cube roots of unity are 
1, — h + hVZ-h — J — Jy'S.i. 

82. Values of cos n0 and sin no in powers of sin o 
and cos fl. 

If we take the equation 

cos nd -\- i sin nd = (cos -\- i sin 0) n 

and expand the right-hand member by the binomial theorem, 

we have 

cos nO -f- i sin n0 = cos" -f »t cos" -1 sin 

n(n — 1) , costt _ 2 ^ gin2 ^ , w(n — l)(n — 2) . 3 3 . 
Li Li 

+ n(n-l)(n--2)(n-3) ., ^ , ^ Q + etc . 

if we introduce the values of P, i 3 , etc., this becomes 
cos n0 -\- i sin = 

/ C os« — n ( n ~ *) cos"- 2 sin 2 
V Li 

n(n-l)(n-2)(n-3) ^ , ^ , _ ^ \ 

L± / 

+ i /n cos"- 1 sin — flfo — 1)0 — 2) cosn _ 3 ^ gin3 ^ _^_ etc \ 

Hence, equating the real and imaginary parts on the two sides 
of the equation, 

cos n0 = cos" — n ^ n ^ cos"- 2 sin 2 

\A 

+ n( n-l)(n--2)(n-3) ^ , gin4 , _ ^ (n3) 

ft 10* 



114 



PLANE TRIGONOMETRY. 



sin nd = n cos n_1 sin 

n (n — l)(n — 2) 

13 



cos"- 3 sin 3 + etc. (114) 



83. Exponential Values of cos ± i sin 0. Expan- 
sion of sin and cos in powers of o. 

One of the most important constants in mathematics is that 
which is known as e. It may be denned by the series 

_^ 

14 



c = i + i +— !— + — 



etc. 



The value of e, as determined by this series, is 2.71828182 
It is proved in most of the text-books of algebra that 



1 + s 



X z X 3 



11 



etc., 



and that this series is convergent for all values of x. (For 
example, see Wells's College Algebra, §§ 521 to, 523, and §465; 
Wentworth's College Algebra, § 392 ; and Chrystal's Algebra, 
Part II., Chap. XXVIII. and Chap. XXIX., §§ 18-20.) By the 
use of this series we have 



e iB 



= 1 + 10 

-C-i 



+ 



& 



W 



l± 



Li ' H 



etc. 



Similarly e~ ie = ( 1 



2 



\1 



[1 



+ etc. 



etc. 



Li 



etc. 



i 



11 + ^. 



etc. 



(115) 



If in (113) and (114) the substitution n0 = a be made, and 
therefore n = — , the results are 



cos a = cos H — 



e \ e ) 



Li 



cos»- 2 sin' 



etc. 



DE MOIVRE'S THEOREM, ETC. 115 



^(f- 1 )^- 2 ) 



sin a = — cos' 1-1 sin — ^— - cos" -3 6 sin 3 

|_3 

-fete. 

These may then be written in the form 

„ a (a — 0) - _ sin 2 . . 

cos a = eos» ^ - cos" -2 — 4- etc. 

[2 # 2 

Wj sin0 a(a — 0)U — 20) ..sin 3 /? . , 

sin a = a cos"- 1 ^ ) cos"" 3 0— 4- etc. 

Now, let ns assume that n increases without limit, while a 
remains constant, then approaches the limit 0. Under these 

circumstances — — approaches the limit 1 (see §44), and cos 

approaches the limit 1. Hence, passing to the limits, we have 
a 2 , a* a 6 



-i a 2 . a* a 6 . , 

cos a = 1 ■ \- etc. 



a° . a a 1 . , 
sin a = a \- etc. 

n la \a 



(116) 



See Chrystal's Algebra, Part II. pp. 256-258. 

Comparison of (116) with (115) reveals at once the following 
relations : 

e ie = cos 
e -i6 _ cos 



e + imaeA (n7) 

— i sin 0.) J 



EXERCISES. 

1. If the equations in (117) be added and subtracted we have for 
the results 

2 cos d = e&-\- e-M and 2i sin 6 = ei — e~ i9 , 

or cos 6 = ' — i sin 6 = — — . 

2 2 

2. From the results of Exercise 1, many of the formulae of 
trigonometry can be proved very readily. Thus 

COS 90 = em + e ~ 2i9 = 2e2W + 2e ~ 2W 

2 4 

_ e 2i9 -\-2-\- e-w , eW — 2 + e-™ 

4 4 

__ (eM-^e^y / eio — e-je y 

= cos 2 + i 2 sin 2 = cos 2 d — sin 2 0. 



116 PLANE TRIGONOMETRY. 

84. The Hyperbolic Functions. It is found that the 

. e% _i_ g-x gx g-x 

functions — -E and present certain striking analo- 

gies to the trigonometric functions whose exponential values 
are given in Exercise 1, above. The first of these is called 
the hyperbolic cosine of x (cosh x), and the second is called 
the hyperbolic sine of x (sink x). Thus we have 

pX _|_ p-X pX p-X 

cosh x = ^ smh x = - — — ^— . (1 18) 

The names of these functions are derived from a certain rela- 
tion they have with a curve called the hyperbola, about which 
the student will learn in the study of analytic geometry. 

Following the analogy of the trigonometric functions, we 
have 

sinh x , , e x — er x /han 

= tanh#=- , (119) 



cosh x e x -f- er x 

which is called the hyperbolic tangent of x. The reciprocals 
of tanh x, cosh x, and sinh x are called respectively the hyper- 
bolic cotangent (cotli), hyperbolic secant (sech), and hyperbolic 
cosecant (csch) of x. 

If in the results of Exercise 1, above, id be substituted for 
0, the following results are obtained : 

cos id = ! i sin id = , 

2 2 ' 

i sin id = — sinh d, 

or sin id = i sinh 0, \ n9Q\ 

and cos id = cosh 0. ) 

85. The relations between the different hyperbolic func- 
tions are in general analogous to, and sometimes identical 
with, the corresponding relations between the trigonometric 
functions. Thus 

i (& _|_ e~*y— I {e x — er x J = 1, 

cosh 2 x — sinh 2 x = 1. 

3 2x e -2x px _|_ p-x e x e -x 



Again, sinh 2x 



2 2 2' 

sinh 2x = 2 sinh x cosh x. 



DE MOIVRE's THEOREM, ETC. 117 

If the sign of x be changed in (118), we have 
cosh ( — x) = ~ — — COfl h x, 

u 

sinh ( — x) = ; = — sinn x. 

We have also 

sinh (x -\- y) = sinh x cosh y -\- cosh x sinh y, 

because 

sinh x cosh y = \- (e x — e~ x ) (& -J- e~y) 

= l (e x +y — e~ x +'J -j- e x ~v — e~ x ^y), 
cosh x sinh y = \- (e x -f- e~ x ) (& — erv) 

= i (e x +u -f- g-'+y — e- r -z/ — e~ x -y), 

adding sinh x cosh y -j- cosh # sinh y = — ^^ 

= sinh (x + y). 
Similarly 

sinh (.r — y) = sinh x cosh y — cosh x sinh 2/, 

cosh (x -j- y) = cosh # cosh y -j- sinh .r sinh y, 

cosh (# — y) = cosh x cosh y — sinh x sinh y. 

The relations given in (120) can often be conveniently applied, 

as in the following reduction : 

sin (a -J- ifi) = sin a cos ifl -f- cos a sin ip 

= sin a cosh /5 -f- i cos a sinh ft 

sin a -j- 1 cos a. 



2 

Thus the expression sin (a -f- ?'/3) is reduced to the typical 
complex form. 

86. The inverse notation for the hyperbolic functions is 
the same as that for the trigonometric functions. (See § 34.) 
Thus, since sinh x = i (e x — e~ x ), 

we have also x = sinh- 1 i (e x — e~ x ). 

The following relation is important : 
Since I (e w — er w ) = sinh w, 

and i (e w -\- e~ w ) = cosh id, 

adding e w = sinh w -j- cosh id, 



118 PLANE TRIGONOMETRY. 

w = log e (sinh w -\- cosh w) ; 
or, since cosh w = yl-\- sinh 2 w 



w = log e (sinh w -f- l/l + sinh 2 w). 
Let sinh w — X, then 20 = sinh" 1 #. 

Substituting these values we have 

sinh- 1 x = log e {pc + l/l + x 2 ). (121) 

The student may prove, in a similar manner, the two following 
formulae : 

cosh- 1 x = log e (x + Vx 2 — 1). (122) 

tanh-i x = J log e *-+j^. (123) 

1 — x 

EXAMPLES. 
1. Find the five fifth roots of — 1 ; the four fourth roots of +- 1. 

2. Find the cube roots of ~- (—1 — iy 3). 

A 

3. If A XJ A 2 , A v A± equal cos a x -f i sin a v etc., show that 
^^4 2 + ^3^4 = 2 COS (<*! + c 2 — « 3 — a 4 ) [cos («! + a 2 + a 3 -f a 4 ) 

4- i sin (a : + a 2 4 a 3 4. a 4 )]. 

4. Also that 

1 sin ( « 1 4- a 2 + a 3 + a J — i COS (Qj -f a 2 4- a 3 -f a 4 ) 

^4^2 — ^l 3 vl 4 2 Sin (fi x 4- a 2 — a 3 — a 4 ) 

5. Also that 

1 cos ((?! -f a 2 -f a 3 -f a 4 ) — 2 sin (^ 4 a 2 + a s + "4) 

(A x 4- -^2) (^3 + -4J 4 COS (a x — a 2 ) COS (a s — a 4 ) 

[See Rev. J. B. Lock's Higher Trigonometry, p. 15, Exs. 1, 2, 
and 4.] 

6. From the results of \ 82 find the expansion of sin nd and 
cos nd 1 in terms of sin 6 and cos 0, for all integral values of n up 
ton = 10. 

7. Verify the following identities by using the exponential 
values of sin 6 and cos 6 given in Exercise 1, p. 115 : 

(1) C os 2 B + sin 2 6 = 1. (5) 2 sin 2 6 = 1 — cos 20. 

(2) sin = —sin (—6). (6) 2 cos 2 = 1 + cos 26. 

(3) cos = cos (— 0). (7) sin 30 = 3 sin — 4 sin 3 ft 

(4) sin 20 = 2 sin cos 6. (8) cos 30 = 4 cos 3 (9 — 3 cos 6. 



DE MOIVRE'S THEOREM, ETC. 119 

Any of the other formulae of trigonometry may be similarly 
verified. 

8. Prove the following relations between the hyperbolic func- 
tions : 

(1) tanh 2 x -\- seen 2 x = 1. (3) sinh 3# = 3 sinh a; -f- 4 sinh 3 #. 

(2) cotb 2 x + csch 2 x = 1. (4) cosh Sx = 4 cosh 3 x — 3 cosh a;. 

(5) tanh (s + y^ ^' + ' f '■■''• 
1 -j- tanh x tanh y 

9. Derive the formulae for the sum and difference of sinh x and 
sinh y, and of cosh x and cosh y. 



x 



10. Prove that sinh- 1 x = cosh- 1 V\ -f- x' 2 = tanh- 1 

l/l+z 2 

11. Prove that tanh- 1 x + tanh- 1 y = tanh- 1 g +y . 

1 + ^ 

12. Prove that cos (a + i(3) = .}- (e/ 3 -f- e-P) cos a — h (c£ — e-/ 3 ) sin a. 



PART II. 



SPHERICAL TRIGONOMETRY, 



INTKODUCTIOlSr. 

87. Spherical Trigonometry treats of the solution of 
spherical triangles. 

A spherical triangle is a portion of the surface of a sphere 
bounded by three arcs of great circles of the sphere. 

Before beginning the study of spherical trigonometry, the 
student will do well to go over in review those portions of 
solid geometry which treat of the sphere and the spherical 
triangle. We learn from the geometry of the spherical triangle 
that it can be determined completely when three parts are 
given, and we do not have to make an exception of the three 
angles as in the case of plane triangles. Since the sum of the 
angles of a spherical triangle is greater than two and less than 
six right angles, it is possible to have triangles with one, two, 
or three right angles. Hence we have the following definition : 

I. Spherical triangles having one, two, or three right angles 
are called respectively spherical right triangles, birectangular 
triangles, and trirectangular triangles. 

It is also possible that one, two, or three sides may be quad- 
rants. Hence we have the definition : 

II. If one, two, or three sides of a triangle are quadrants, 
the triangle is called respectively a quadrantal, biquadrantal, 
or triquadrantal triangle. 

It is shown in geometry that a trirectangular triangle is 
also triquadrantal, and is equal to one-eighth of the surface of 
the sphere. 
120 



INTRODUCTION. 



121 



Fig. 32. 



Each of the parts of the spherical triangle will be taken less 
than 180°. 

88. A spherical triangle and a trihedral angle are essen- 
tially the same thing. The sides of the triangle correspond 
to the face angles of the trihedral angle, and the angles of the 
triangle to the dihedral angles of the trihedral angle. 

A figure will make this clear. 
Let ABC be any triangle upon 
the surface of the sphere whose 
centre is at O. Let the planes 
of the sides of the triangle be 
constructed, meeting, two and 
two, in the lines OA, OB, and 
OC These three planes will 
form a trihedral angle at 0. 
The face angles of this trihedral 
angle, A OB, BOC, and CO A, are measured by c, a, and b re- 
spectively, and the angles of the triangle A, B, and C are the 
same as the dihedral angles whose edges are OA, OB, and OC 
respectively. 




U 



CHAPTEE IX. 



GENERAL FORMULAE. 



Fig. 33. 



89. In any spherical triangle the sines of the sides are propor- 
tional to the sines of the opposite angles. 

In Fig. 33 let ABC be any spherical triangle, the centre 
of the sphere, and OA, OB, and 
OC the edges of the dihedral 
angles formed by the planes 
of the sides of the triangle. 
From B', any point of OB, 
draw B'P perpendicular to 
AOC, and through i?'P pass two 
planes perpendicular to OA and 
OC respectively. The first of 
these will intersect the planes 
AOC and AOB in the lines PA' 
and B'A', both perpendicular to OA, and the second will inter- 
sect AOC and BOC in PC and B'C, both perpendicular to OC. 

In the right triangles A'PB' and C'PB' we have 




sin A' 



sin A 



Hence 
From th< 



B'P 
A'B" 
B'P 
A'B" 
sin A 
sin C 



sin C : 
sin C - 

MB'' 



B'P 
B'C 
B'P 

B'C' 



riodit triano-les A' OB' and COB' we have 



sin A' OB' 



sin c 



A'B 



Hence 
122 



OB' 7 

A'B' 

OB' 

sin a 



sin COB' = 



FC 
OB' 



sin c 



sin a 

B'C 
A'B'' 



B'C 
OB'' 



by § 88 

to 

by § 88 

W) 



GENERAL FORMULAE. 



123 



(124) 



Therefore, from (a) and (/3), 

sin a sin A 

sin c sin 
Thus the same thing may be proved for each pair of sides 
and angles, and we may write 

sin a sin b sin c 

sin A sin B sin C' 

or, in the form of three separate equations, 

sin a sin B = sin b sin A, 

sin b sin C = sin c sin j5, 

sin c sin J. = sin a sin C 

It should be observed that the figure can be drawn equally 
well when one of the angles, as A, is obtuse. The foot of 
the perpendicular B'P will then lie outside the space enclosed 
by A OC, but the demonstration applied to such a figure is 
precisely like that above, so that the theorem there established 
is perfectly general. This remark applies also to that which 
follows. 

90. In any spherical triangle the cosine of each side is equal to 
the product of the cosines of the other two sides phis the product 
of the sines of these sides and the cosine of their included angle. 



Fig. 34 is constructed in the 
same manner as Fig. 33, with 
the addition of A'M drawn in 
the plane AOC perpendicular 
to OC, and FN parallel to OC. 
OC OM , MC 



Fig. 34. 



OB' 



^ OB' 

. pjst . 

OB' h OB' ' 



OB' 
OM 



w 



but 



OC 



OB' 



= cos a, 




OM 



OA' 



= — ^— X ^^- = cos b cos c, 
OB' OA' OB' 



124 



SPHERICAL TRIGONOMETRY. 



, PN _PN A'P A'B' 
and '0B'~JJP X A F B' X OW 



sin PA'N cos A sin c 7 



= sin b sin c cos A, 
since PA'JY= AOC—b (sides perpendicular each to each). 
Substituting in (a) these values of , ,and , we have 

cos a = cos b cos c -j- sin b sin c cos A. } 
Also, cos b = cos c cos a -j- si n £ sin <2 cos B, j- (125) 

cos c = cos a cos b -j- sin a sin 6 cos (7. J 

These formulae will furnish 
a theoretical solution of a tri- Fig. Str- 

angle when two sides and the 
included angle are given, or 
when the three sides are given; 
but they are not well adapted 
for logarithmic computation. 
For this reason they are seldom 
used in this form. Other for- 
mulas derived more or less 
directly from them are found 
to be more useful in practice. 

The most convenient methods for the complete solution of 
triangles are given in Chap. XI. When, however, two sides 
of a triangle and the included angle are given, and the third 
side only is required, we can use (125) by means of the device 
shown in the following exercise. 

Given A = 73° 25', b = 38° 12', c = 49° 37'. To find a. 

In the formula 

cos a = cos b cos c + sin b sin c cos A, 
let k sin = cos &, (a) 

and h cos 6 = sin b cos A. (/?) 

Substituting these values of cos b and sin 5 cos A, 
we have cos a = & sin 6 cos c + k cos 6 sin c, 

or, by (48) cos a = k sin (0 + c) ; 

where the auxiliary quantities k and 6 are found from (a) and 
(P) as in § 72. 
Thus tan 6 




cos 6 



cot b 



sin 6 cos A cos .4' 



GENERAL FORMULAE. 



125 



and 



sin 



sin b cos A 

cos 



The logarithmic work may be arranged as follows : 



loa; cot b 



+ c 

+ c 



= .10407 
= 9.45547 
= .64860 
= 77° 20' 31" 
= 257° 20' 31" 
= 126° 57' 31" 
= 306° 57' 31" 



log cos b 
log sin 6 
log & 



9.89534 
= (±) 9.98932 



= (±) 9.90602 
log sin (0 + c) = (±) 9.90258 
log cos a 9.80860 

a = 49° 56' 28" 



It should be noted that while 8 has two values, and hence both 
k and sin (0 + c) have two equal values, positive and negative, a 
has only one value, since the upper signs of k and sin (6 -f c) must 
be taken together, and also the lower signs. 

EXAMPLES. 
1. Given B = 51° 16', a = 38°, c = 74°. Find 6. 
2. Given (7= 102°, a = 85°, 6 = 63°. Find c. 

91. Polar Triangles. We know from geometry that if 
ABC and A'B'C be two polar triangles, 

A = 180° — a ',a= 180° — A', 

B = 180° — 6', 6 = 180° — B' : 

C = 180° — c', c = 180° — (7'. 
By taking a formula that applies 
to any spherical triangle and sub- 
stituting the above values for the 
various quantities in the formula, we 
can derive a new formula of a similar 
form to the old, but having angles 
where the original had sides, and vice 
versa. It is obvious that the new for- 
mula must be just as general as the 
old. This process is called " applying 
the formula to the polar triangle." 

If we take (124) and apply it to the polar triangle, we have 
sin (180°— A') sin (180°— &') = sin (180°— B') sin (180°— a'), 
or sin A f sin V = sin B' sin a\ 

n* 




126 



SPHERICAL TRIGONOMETRY. 



which is the same as that with which we started. Hence 
nothing has been gained. 

92. If we make the same substitution in (125), we have 
cos (180° — A') = cos (180° — B') cos (180° — C) 
-f sin (180° — B') sin (180° — C) cos (180° — a'), 
or — cos A' = cos B' cos C — sin B! sin C cos a'. by (42) 
which is a new formula, giving the cosine of an angle in terms 
of the other angles and their included side. Thus we can 
write 

cos A = — cos B cos C -f- sin B sin C cos a, \ 
cos B = — cos C cos A -(- sin C sin A cos b, [■ (126) 
cos G= — cos A cos B -f- sin A sin B cos c. J 
Formulae (126) may be used, in the same manner as (125), 
to find the unknown angle when two angles and the included 
side are given. Thus in 

cos A = — cos B cos C -J-- sin B sin C cos a 



substitute 




k sin 6 = — cos B : 
k cos 6 = sin B cos a. 


Then 




, Q cot B 
tan $z= , 

cos a 




k = - 


cos B 7 sin B cos a 

: , or k = — , 

sm d cos 6 


and cos A - 


= k sin 6 cos C -f- k cos sin (7, 


or cos A ■ 


= k sin (0 -|- (7). 






EXAMPLES. 


1. Given a 


= 74°, 


.5 = 103°, 0= Fig. 36. 


61°. Find A. 




B 


2. Given c 


= 38°, 


A =51°, ,B= j//\ 


69°. Find C. 







93.* Let A£(7 be any spheri- 
cal triangle, and draw an arc BD 
from B to any point of A C. Then 
we have from the triangles ABD 
and BCD by (125), 




* This demonstration was suggested by Mr. George Hervey Hallett. 



GENEBAL FORMULAE. 



127 



0) 
(JO 



cos BD = cos a cos CD -j- sin a sin CD cos (7, 
cos BD = cos c cos AD -f sin c sin AD cos A. 
Since AD = b— CD, we have by (51) and (49) 

cos AD = cos b cos CD -(- sin 5 sin CD, 
sin AD = sin 6 cos CD — cos b sin CD. 
If we substitute these values in (/3) we have 
cos BD = cos 6 cos c cos CD -(- sin b cos c sin CD 

-f- sin 6 sin c cos J. cos CD — cos b sin c cos A sin CD 
= (cos b cos c + sin & sin c cos A) cos CD 

-f- (sin b cos c — cos b sin c cos J.) sin CD) 
but cos 6 cos c -f sin & sin c cos A = cos a, by (125) 

. • . cos BD = cos a cos CD 

-j- (sin b cos c — cos b sin c cos A) sin CD. (p) 
Hence, equating the values of cos BD in (a) and (^) and col- 
lecting terms, we have 

sin a cos C sin CD = (sin 6 cos c — cos b sin c cos J.) sin CD, 
or sin a cos C= sin 6 cos c — cos b sin c cos A. 

By transposing the letters five other formulae similar to this 
may be written. The whole group of six is given below. 

sin a cos C = sin b cos c — cos b sin c cos A, 
sin a cos B = sin c cos 6 — cos c sin 6 cos A, 
sin b cos A = sin c cos a — cos c sin a cos B, 
sin 6 cos C = sin a cos c — cos a sin c cos B, 
sin c cos A = sin 6 cos a — cos b sin a cos C, 
sin c cos B = sin a cos b — cos a sin b cos C. 

94. If we apply (127) to the polar triangle, we derive the 
following: : 



(127) 



sin A cos b = sin C cos B -\- cos C sin B cos a, \ 
sin A cos c = sin B cos C -f- cos B sin C cos a 
sin i? cos c = sin JL cos C -j- cos J. sin C cos 6 
sin 5 cos a = sin C cos J_ -j- cos C sin A cos 6, 
sin C cos a = sin B cos J. -j- cos B sin .A cos c, 
sin C cos 6 = sin A cos 5 -j- cos A sin i? cos c. 



(12S) 



(129) 



128 SPHERICAL TRIGONOMETRY. 

95. If we now take the first formula of (127) and substi- 
tute in it sin a = , from (124), we have 

sin C 

sin c sin A cos G • -, i . A 

== sin o cos c — cos b sin c cos A, 

sin G 

cos O 
or, dividing through by sin c, and writing cot G for — , this 

reduces to 

sin A cot C = sin b cot c — cos b cos A. 

Eearranging the terms, we have finally cot c sin b = cos b 
cos A -\- sin A cot G This, with the five similar formulae 
obtained by transposing the letters, are given below, 
cot c sin b = cos b cos A -f sin A cot G, 
cot c sin a = cos a cos B -j- sin B cot (7, 
cot & sin a = cos a cos G -f- sin (7 cot B, 
cot 6 sin c = cos c cos A -J- sin J. cot 2?, 
cot a sin c = cos c cos ^ -j- sin B cot ^1, 
cot a sin 5 — cos b cos (7 -)- sin G cot A 

96. The foregoing sets of formulae, while very useful for 
many purposes, are not given, if we except (124), in a form 
convenient for logarithmic computation. Thus (125) will give 
the three angles if the sides are known, and (126) will give 
the three sides if the angles are known ; but to reduce them 
to a form that will be convenient in practice they must be 
considerably modified. 

From (125) we have 

cos a — cos b cos c /10 n N 

cos A = — — . (130) 

sin b sm c 

Subtracting both sides from unity, and remembering that, 
by (57), 1 — cos A = 2 sin 2 %A, 

we have 

• 2 i a sin 6 sin c + cos b cos c — cos a 

A Sin 2" A. — — ; ; " , 

sin b sin c 

. o , A cos (b — c) — cos a -, TT /F .-,. 

or sm 2 \A = ^— - — - L —. . by (51) 

2 sin b sin c 

In (63), let <p = a and d = b — c. 



GENERAL FORMULAE. 



129 



We then have 
cos (b — c) — cos a = 2 sin £ (a — b -j- c) sin 2 (a -f- b — c). 



Now, let 



2s = a -f b + c : 
__26 = a — 6-|-c: 



W 



(s — 6) = i(a — 6 + c); 

also, (s — c) = * (a 4-6 — c). 

Hence cos (b — c) — cos a = 2 sin (s — U) sin (s — c). 

Therefore we have 

S i n 2 1^ = sin (s — b) sin (5 — c) ^ 
sin b sin c 
sin (s — c) sin (s — a) 

sin c sin a 

sin (s — a) sin (s — b) 

sin a sin 6 

97. If both members of (130) be added to unity, remem- 
bering that, by (58), 

1 + cos A = 2 cos 2 M, 
we have 



Also, 



sin 2 \B 
sin 2 hC 



(131) 



2 cos 2 IA 



sin 6 sin c — cos b cos c -j- cos a 
sin b sin c 



or 



cos 2 }A = cosa-cos(5 + g ) t 
2 sin b sin c 
p = b -\- c and 6 = a. 



by (50) 



In (63), let 
We then have 

cos 2 U = 2 sin * (& + f + fl)sin l(b + c — d) 



As before, let 
then 



2 sin b sin c 
s = i (a + 6 + c), 



I, ., i i sins sin (5 — a) y 
and we have cos 2 \A = : ^ *-. 

sin b sin c 



Also, 



COS" 



id sin s sin (s — b) 



sin c sin « 



(132) 



COB , i(7= Binsma_(s--£) 

sin a sin 6 



130 



SPHERICAL TRIGONOMETRY. 



98. By dividing each of (131) by the corresponding one 
of (132), we find 

tan 2 U = sin (8-6) sin (8 -c) 
sin s sin (s — a) 

tan' \B = Bi°(»-c)ain(«-a) 

sin s sin (s — 6) 

tan 2 i^ sin (g — a) sin (s — 6) 
sin s sin (s — c) 

If we use here the same device that was employed in the 
corresponding case of plane triangles, § 58, and take as an 
auxiliary quantity 



tan 2 r 



sin (s — a) sin (s — b) sin (s — c) 



sin s 



(134) 



we have 
tan 2 \A 



sin (s — a) sin (s — b) sin (s — c) _ tan 2 r 



sin s sin 2 (s — a) 
tan r 



tan \A 
tan Ji? 



sin (5 — a) 

tan r 
sin (s — 6)' 



tan J C = — 



tan r 



sin 2 (s — a)' 



(135) 



sin (s — c) 

Any one of the sets of formulae, (131), (132), (133), or (134) 
and (135), may be used to find the three angles of a triangle 
when its sides are known. Next will be derived similar 
formulae for finding the sides when the angles are known. 

99. From (126) we have 

COS B COS C 4- COS A s-ioa\ 

cos a = : — T^ . (136) 

sin B sm C 

Subtracting both members of this from unity, we have, by (57), 
sin B sin C — cos B cos C — cos A 



2 sin 2 \a = 



or 



sin B sin C 

cos (B + C) + cos A 
2 sin B sin G 



by (50) 



GENERAL FORMULAE. 

In (62), let <p = B -f C, = A. 

Then 

8in2 la = - 2 coBiQl + *+C)cosj ( g+C-,Q 



131 



Let 
then 



Also, 



2 sin B sin G T 

S=i(4 + .B+C), 

S— A = i(B+ C—A): 

8 j n * i a = _ *os S cos (S- A) 
sin i? sin C 



sin 2 2 6 



cos S cos ($ — 5) 
sin G sin J. 

cos £cos (S—C) 



(137) 



sin J. sin B 

100. If both members of (136) be added to unity, we have 
9 . i sin 5 sin C + cos J? cos (7+ cos J. 

-j COS 2^ — 



sin B sin C 

cos 2 *a = cos (^-(7) + cos A 
2 sin .£ sin G 

<p = A and = B — G. 



by (58) 
by (51) 



In (62), let 
We then have 

cos 2 \a = 2 cos j Q* - * + O cos j Ql + JB- g) 
2 sin jB sin (7 

As before, let S = I (A -f .8 + C), 

then 

(£— 5) = J (A — -5 + C), and (S— C) = } (A -f £ — G) ; 

cos^a^ cos ^-:^ COS ^- a) . 1 

sin B sin C 



Also, cos' 16 = <**(S-C)<x*(B-A) 

sin (7 sin A 

cos 2 *c = cos(S-,i)cos(£-i?) 
sin A sin 5 



(138) 



101. By dividing each of (137) by the corresponding one 
of (138), we obtain 



132 



SPHERICAL TRIGONOMETRY. 



tan 2 ha 



tan 2 \c 



cos ff cos (S — A) 



cos (# 


--B) 


COS 


es - 


cy 


cos 


£ COS 


OS- 


-B) 




cos (#- 


-C) 


COS 


(S- 


*1 


COS 


#COS 


OS- 


-C) 





(139) 



cos (S— A) cos (S—B) 

It will have been observed that in each of the equations 
(137) and (139) the right-hand members have the negative 
sign, while the left-hand members are perfect squares. This 
inconsistency, however, is only in appearance. We know from 
geometry that 2S > 180° and < 540° : therefore S > 90° and 
< 270°. In other words, cos S is always negative, and hence 
— cos S is always positive. 

If we employ in connection with (139) an auxiliary quantity 

— cos S 



tan 2 B 

we have 

tan 2 la 



cos (S— A) cos (S—B) cos (£- 

— cos S cos 2 (S — A) 

cos (S — A) cos (3 — B) cos (S- 

= tan 2 B cos 2 (S—A). 

tan la = tan B cos (S — A), 

tan i6 = tan B cos (S — B), 

tan lc — tan B cos (S — C). 



G) 



cy 



(140) 



(141) 



102. Napier's Analogies. If we multiply together 
the first two equations of (133) and reduce the result to its 
simplest form, we have 



tan I A tan i B 



sin (s — c) 



sin s 

^Regarding this as a proportion, we have, by composition and 

division, 

1 -f- tan \A tan \B sin s -\- sin (s — c) 

1 — tan \A tan \B sin s — sin (s — c) 

1 + tan \A tan \B = cos \ (A — B) 
1 — tan \A tan \B cos i (A -\- i?)' 



But 



by (71) 



GENERAL FORMULAE. 133 

and sin s -f- sin (s — c) tan j (2s — e) 

sin s — sin (s — c) tan *c by ( W ) 

= tan -»- (a-j-b) . 
tan £c 
cos £(4 — ■#) = tan J ( rt -f- ft) 
cos I- (J. 4- B) tan Jc ' 

tan J (« + ft) = °^ff |) tan ^ (142) 

cos J (vl -f B) ' ^ ' 

which is the first of Napier's Analogies. 

103. Divide the first of (133) by the second, and reduce 
We have 

tan \A ___ sin (s — ft) 
tan %B ~ sin (s — a) 
By division and composition, 

tan \A -— tan jB _ sin (s — ft) — sin (s — a^ 
tan IA + tan %B sin (s — ft) -f sin (5 IT^y 
But tan M — tan ^ __ sin i (A — B) 

tan |4 -f tan \B sin J (4 -f- £)' by (70 ) 

and Bin (s — b)-- sm (s — a) tan \ (a — ft) 

sin (s _ ft) + sin (s - «) tan J (2s — a—*) J (64) 

__ tan_J_^-_ft) 
tan \c 
sin j (4 — £) = tan| (a — ft) 
sin J (^ + B) ' tan Jc ' 

or tan * (a- ft) = !iB_LOtzzA) tan ^ n^ 

sin * (^ + *) an 2 ^' (143 ) 

which is the second of Napier's Analogies. 

104. In a precisely similar way, hy using the first two 
equations of (139), or else by applying (142) and (143) to the 
polar triangle, we obtain 



12 



134 SPHERICAL TRIGONOMETRY. 

The student will find the work of deriving these formulae 
by both these methods profitable exercise. 

105. Gauss's or Delambre's Formulae. 

By (48) sin J (A + B) = sin \A cos \B -J- cos \A sin \B. 
In this equation substitute for sin \A, cos %B, etc., their 
values as given in (131) and (132), we then have 

sin i (A + B) = sin(s- j)Jju: s S in(s- C ) 
sin c * sin a sin b 



. sin (s — a) I sin s sin (s — c) 
sin c \ sin a sin b 



sin (s — b) -f- sin (s — a) I sin s sin (s — c) 

sin c \ sin a sin 6 

low, by (60) 
sin (s — 6) -f sin (s — a) = 2 sin J (2s — 6 — a) cos \ (a — b) 

= 2 sin \c cos \ (a — 6), 
and by (54) sin c = 2 sin Jc cos Jc ; 

while the radical is the value of cos i C in terms of the sides. 

• i / a i ©x 2 sin |c cos $ (a — b) iri 

sin J (i + 5) = =; ^ * cos J (7, 

2 sin %c cos ?c 

or cos \c sin J (A -f- B) = cos J (a — b) cos ^C, 

which is the first of Gauss's formulae. 

Starting now with 

cos i (A -f B) = cos \A cos \B — sin \A sin %B, 

we go through the same steps as above ; thus 

, , A . t, n sins /sinfs — a) sin (s — 6) 

cos J (A A- B) = — — \ * — ; — J . - \ J - 

sm c \ sin a sm 6 

sin (s — c) I sin (s — a) sin (s — 6) 
sin c \ sin a sin 6 

by (131) = 8i°«-Bin(«-c) gin |0 

sine 

by (61) = 2eo 8 K2^-c)sini C sin } ^ 

2 sin Je cos £c 

= cob Ha + ») gtnt g 

COS ^c 



GENERAL FORMULAE. 



135 



cos $c cos i (A + B) = cos | (a -f b) sin * C, 
winch is the second of Gauss's formulae. 
Again, 

sin £ (A — B) = sin |A cos IB — cos \A sin \B 
- sin (s — b) — sin (s — a) QQs ^ 



sin c 



as above for sin i (A -f- J?) 
sin £ (J. 

_ sin £ (a 



B) = 2co8icBinlO»-t) C os |C, 

2 sin ic cos ^c 



'') 



cos iC. 



sin ic 

sin \c sin £ (A — B) = sin J (a — b) cos JO, 
which is the third of Gauss's formulae. 
Finally, 

cos i {A — B) = cos \A cos \B -f- sin £A sin \B 



sin s + sin (s — c) sin ^ 



sin c 

2 sin U« 4- J) cos *c . 2 n 

= =-5> — -3- — ^ — — =- sin £ C 

2 sin \c cos *c 

= sinj_(a + 6) siniC , 

sin \c 

sin |c cos J (A — B) = sin I (a -f 6) sin J 0, 

which is the fourth of Gauss's equations. 

For convenience in reference these four results are placed 

together. 

cos ic sin % (A -{- B) = cos | (a — b) cos J(7. 

cos ic cos J (J. + B) = cos J (a -f 6) sin i C 

sin Jc sin £ ( A — B) = sin } (a — b) cos § (7. 

sin 2^ cos | (A — 5) = sin £ (a -f- 6) sin J (7. 

These equations may be used to derive Napier's Analogies, 

Thus if we divide the fourth by the second we have 

sin \e cos J (A — B) sin $ (a -\- b) sin iC 

cos lc cos i (i -f 5) cos i (a-\- b) sin JC' 



(146) 



or 



tan Jc 



cos | (A 



Q = tan i (a + 6), 



cos i (A -f _B) 
which is the first of Napier's Analogies 



136 SPHERICAL TRIGONOMETRY. 

The third of Gauss's Equations divided by the first gives 
the second of Napier's Analogies ; the first of Gauss's Equa- 
tions divided by the second gives the third of Napier's Analo- 
gies ; and the third of Gauss's Equations divided by the fourth 
gives the fourth of Napier's Analogies. 

106. EXERCISES. 

1. Let n 2 = sin s sin (s — a) sin (s — b) sin (s — c). 

To prove 4n 2 = 1 — cos 2 a — cos 2 b — cos 2 e + 2 cos a cos b cos c. 
By (63) 2 sin s sin (s — a) = cos a — cos (2s — a), 

= cos a — cos (6 -f c) ; 
also, 2 sin (s — b) sin (s — c) = cos (b — c) — cos (2s — b — c), 

= cos (b — c) — cos a. 
Multiplying these two equations, we have 
4n 2 = 4 sin s sin (s — a) sin (s — b) sin (s — c) 

= cos a [cos (6 + c) + cos (6 — c)] — cos 2 a — cos (b + c) cos (b — c) . 
Reducing this by (62), (50), and (51), we have 

An 2 = 2 cos a cos 6 cos c — cos 2 a -^ cos 2 b cos 2 c + sin 2 b sin 2 c, 
or, since sin 2 b sin 2 c = (1 — cos 2 b) (1 — cos 2 c), 

we have, after substituting and reducing, 

4n 2 = 1 — cos 2 a — cos 2 b — cos 2 c + 2 cos a cos b cos c. 
(See Chauvenet's Trigonometry, p. 163.) 

This quantity n 2 is an important auxiliary quantity in spherical 
trigonometry, n is called the sine of the solid angle that the tri- 
angle subtends at the centre of the sphere. 

2. If in a spherical triangle p x , p 2 , and p z represent the three 
perpendiculars from the vertices to the opposite sides a, b, and c 
respectively, show that 

sin a sin p x = sin b sin p 2 = sin c sin p 3 = 2n. 
If we draw a triangle and construct the perpendicular^, we have 

from (124) 

sin p sin b . 

sin C~ sin 90°' 

sin p = sin b sin C. 
Hence we are to prove sin a sin b sin C= 2n. 
Now, by (54) sin C= 2 sin | C cos I C, 



by (131) and (132) = 2 J sin s sin (s-a) sin (s-6) sin (s- c) 



sin 2 a sin 2 6 



GENERAL FORMULiE. 137 

Multiplying by sin a sin b, we have 

sin a sin b sin C— 2 V sin s sin (s — a) sin (s — 6) sin (s — e), 

That sin 6 sin £> 2 and sin c sin p 3 are equal to the same thing can 
be shown by simply interchanging the letters. 



3.* To prove 99lA±^sB = sin (« + 6) . 
1 — cos C sin c 



In any right triangle construct the internal bisector of the angle 

C, and let this bisector make with the side c an angle 6. Then we 

have in the two triangles thus formed, by (126) 

cos 6 -|- cos A cos %C= sin A sin W cos 6, 

—cos 6 -\- cos B cos $C= sin B sin £Ccos a. 

Adding these equations, we have 

(cos A + cos B) cos £ C= (sin ^4 cos 6 + sin i? cos a) sin |C. 

Substituting 

sin A = sina sin C , and sin 1? = sin 6 8in C , 
sin c sin c 

we have 

(cos ^L + cos B) cos £C= 8m ? 8in ^ C (sin a cos b + cos a sin 6), 

sin c 

by (54) and (48) = ***¥><*»& sin (a + b) ; 

sin c 

or, since 2 sin 2 JC= 1 — cos C, (57) we have, after reduction, 

cos ^4 -f- cos i? sin (a -\- b) 

1 — cos C sin c 

4* Show that gojL f4 - cos i? = sin (»-«) _ 
1 — cos C sin c 

This can be done in the same manner as in Exercise 3, by drawing 
the external bisector of the angle C. 

EXAMPLES. 

1. Prove the following relations between the parts of any 
spherical triangle. 

,-.s cot \A cot \B cot \C 



sin (s — a) sin (s — b) sin (s — c) 
(2) tan ^B tan \C= sin [ s ~ a \ 



* See McClelland and Preston's Spherical Trigonometry, Part I., third 
edition, p. 51. 

12* 



138 SPHERICAL TRIGONOMETRY. 

(3) tan J6 tan \c = ~ cos S . 
w 2 2 cos{S—A) 

(4) ' 

(5) 

(6) 

(7) 

(8) 

(9) 

cos Jc sin C 



cos J.4 cos hB __ 


_ sin s 


sin tj<7 


sin c 


cos \A sin ^jB _ 


_sin — a) 


cos \C 


sin c 


sin J^4 sin Ji?_ 


_ sin (s — c) 


sin \C 


sin c 


sin Ja sin \b _ 


—cos # 


cos Jc 


sin C 


sin Ja cos \b _ 


_cos(#— .4) 


sin ^c 


sin (7 


cos \a cos J6_ 


_ cos (S—C) 



2. If i? = A + (7 and D be the middle point of b, show that b = 
2,BZ>. 

3. If B be the middle point of 6, show that 

cos c + cos a = 2 cos ^6 cos i?Z>. 

4. If 6 + c = 7T, show that sin 2B + sin 2(7= 0. 

5. In an equilateral triangle show that 

(1) 2 cos A = 1 — tan 2 J a. (2) 2 cos \a sin J^4 = 1. 

(3) sec A = 1 + sec a. 

6. If b = e = 2a, show that esc \A = 4 cos a cos £a. 

7. If a = Jtt, show that cos A -\- cos B cos (7=0. 

8. If c = j7r, and X be any point on c, show that 

cos CX= cos a sin AX-\- cos 6 sin 5X 

9. If the arc of a great circle be drawn in any spherical triangle 
from (7 to the middle point of c, and if this bisector makes an 
angle <$> with c, show that 

cot A — cot B = 2 cos Je cot 0. 

10. In any spherical triangle bisect aiiiD and bisect AD in E. 
Then show that 

cos CE + cos BE =2 cos BD cos BE. 

11. Show that 8?z 3 = sin 2 a sin 2 6 sin 2 c sin ^4 sin B sin (7. (See 
Exercise 2, p. 136.) 

12. Show that -^— = sin a sin 6 sin e sin \A sin Ji? sin J C. 

sin 

13. Show that 

sin A = ■ *» , sin B = *» gin c== 2n 

sin 6 sin c sin c sin a sin a sin o 



GENERAL FORMULA. 139 

14. If N 2 = — cos S cos (S— A) cos (S— B) cos (S— C), show that 
sln a = . **. „, sin b = 2JT sin c 



sin i? sin C sin (7 sin ^4 sin A sin i? 

15. Show that 

4iV 2 = 1 — cos 2 A — cos 2 B — cos 2 (7—2 cos J. cos B cos (7. 

16. If p x , p 2 , p 3 have the same meaning as in Exercise 2, p. 136, 

show that 

sin A sin ^ = sin B sin p 2 = sin C sin p 3 = 2N. 

17. Show that Si^ 3 = sin a sin 6 sin c sin 2 ^4 sin 2 B sin 2 (7. 

18. Show that ra = - — — ^ v D . „ , iV r =- ^V^ — • 

sin ^4 sin ,6 sin C sin a sin 6 sin c 

19. Show that 4niV= sin a sin 6 sin c sin ^4 sin B sin (7. 

20. Prove the following relations between the parts of any 
spherical triangle : 

(1) cos a = cos 6 cos c + sin a sin b sin C cot ^4. 

(2) 2 cos I {a + 6) cos J (a — 6) tan |c 

= sin b cos .4 -f sin a cos i?. 

(3) cos i? -f- cos C _ 2 sin (s — 5) sin (.9 — c ) 

sin (6 -+- c) sin a sin 6 sin c 

(4) tan H-4 — «) tan £ (^ + 6) = tan J {B — 6) tan | {A -f- a). 

(5) cot a = sln 2c eos B ~ siD 26 cos C . 

cos 26 — cos 2c 

(6) (cos 2.4 — cos 2B) — (cos 2a — cos 26) 

= cos 2A cos 26 — cos 2a cos 2B. 



CHAPTEE X. 

THE SOLUTION OF SPHERICAL RIGHT TRIANGLES. 

107. Formulae. The formulae derived in the preceding 
chapter apply to spherical triangles in general. The formulae 
for spherical right triangles may be derived from them by* 
giving to one of the angles the value 90°. Let this angle be 
C. If then we take 
from (124), 

sin a sin C= sin c sin A, and sin b sin C= sin c sin B; 
from (125), cos c = cos a cos b -f- sin a sin b cos C; 

and from (126), cos A = — cos B cos C-\- sin B sin C cos a, 
and cos B = — cos C cos A -\- sin C sin A cos b ; 

and if in each of these we make C= 90°, we have 

sin a = sin c sin A, (147) 

sin b = sin c sin B, (148) 

cos c = cos a cos b, (149) 

cos J. = sin B cos a, (150) 

cos B = sin .A cos 6. (151) 

Additional formulae may be derived as follows : the product of 
(148) and (151) gives sin c sin B sin A cos b = sin b cos B, 
or sin c sin A = tan 6 cot B ; 

. - . from (147), sin a = tan b cot £. (152) 

Hence, also, sin b = tan a cot A. (153) 

The product of (150) and (151) gives 

cos a sin B cos b sin A = cos J. cos 5, 
or cos a cos 6 = cot A cot 5 ; 

. • . from (149), cos c = cot A cot B. (154) 

The product of (148) and (149) gives 

sin c sin J5 cos a cos & = sin b cos c, 
or sin B cos a = tan b cot c; 

.-. from (150), cos A — tan b cot c. (155) 

Hence, also, cos B = tan a cot c. (156) 

140 



SOLUTION OF SPHERICAL RIGHT TRIANGLES. 141 

These ten formulae will solve any case of right triangles. 
They are most easily remembered by the aid of two arbitrary 
rules invented by Napier, which depend upon certain pecu- 
liarities the formula) exhibit when their manner of expression 
is slightly changed. Throughout the ten, in place of c, A, and 
B will be written the complements of those parts, expressed 
as (co. c), (co. A), and (co. B), with the necessary change of 
function to preserve the truth of the equation. Thus, (147) 
will become sin a = cos (co. c) cos (co. A), which is obviously 
the same as the form given above, since cos (co. c) = sin c and 
cos (co. A) = sin A. The formula) are given below as they 
appear when this change is made. They are arranged in two 
groups of five each. 

sin a = cos (co. c) cos (co. A), 
sin b = cos (co. c) cos (co. B), 
sin (co. c) = cos a cos b, 
sin (co. A) = cos (co. B) cos a, 
sin (co. B) = cos (co. A) cos b. 

sin a = tan (co. B) tan b, 

sin b = tan (co. ,4) tan a, 
sin (co. c) = tan (co. A) tan (co. _B), 
sin (co. A) = tan (co. c) tan b, 
sin (co. B) = tan (co. c) tan a. 

It now appears that the sine of each part of the triangle 
(the right angle is, of course, excepted) is expressed in the 
first group in terms of the product 
of the cosines of two other parts, Fig. 37. 

and in the second group in terms 
of the product of the tangents of 
two other parts. Moreover, if the 
figure (Fig. 37) be examined, it will 
appear that in the case of the first 
group the two parts on the right- 
hand side of each equation are none 
of them immediately adjacent to the part on the left. Thus, in 
(147), c and A are each separated from a, the one by B and the 




142 



SPHERICAL TRIGONOMETRY. 



Fig. 37. 




other by b (C is treated as if it had no existence), and in (149) 
a and b are separated from c by B and A respectively ; and 
so for the others. These parts are 
therefore said to be opposite to the 
part on the left. In the second 
group, however, the parts on the 
right of each equation are always 
immediately adjacent to the part 
on the left (treating C as before). 
The part of the triangle on the left 
of each equation is called, for con- 
venience, the middle part. We are now prepared to enunciate 
Napier's rules. In any spherical right triangle — 

I. The sine of the middle part is equal to the product of the 
cosines of the opposite parts. 

II. The sine of the middle part is equal to the product of the 
tangents of the adjacent parts. 

It will assist the memory to observe that the vowel o occurs 
in cosine and opposite, while a occurs in 
tangent and adjacent. It must not be 
forgotten that by the word parts in 
these rales is meant the sides of the 
triangle and the complements of the two 
angles and of the hypothenuse. Fig. 38 
is sometimes a convenient help in de- 
ciding the relative positions of any 
parts under consideration. 

108. Theorems. Before applying Napier's rules to the 
solution of triangles we will establish two theorems : 

I. In any spherical right triangle a side and its opposite angle 
are both > 90° or both < 90°. 

cos B 



Fig. 38. 




From (151), 



sin A = 



cos b 



Now, A is always < 180° : therefore sin A is always positive. 
Hence cos B and cos b must have the same sign ; which proves 
the proposition. 



SOLUTION OF SPHERICAL RIGHT TRIANGLES. 143 

II. If the two sides of a spherical right triangle are both > 
90° or both < 90°, the hypothenuse is < 90° ; but if one side is 

> 90° and the other < 90°, the hypothenuse is > 90°. 
From (149), cos c = cos a cos b. 

If a and b are both > 90° or both << 90°, cos a and cos b have 
the same sign. Therefore cos c is positive, and c is << 90°. 
If, however, a is > 90° and 6 < 90°, or wee versa, cos a and 
cos 6 have opposite signs. Therefore cos c is negative and c is 

> 90°. 

109. Application of Napier's Rules. We now proceed 
to apply Napier's rules to the solution of some right triangles. 
A right triangle can be solved if two of its parts besides the 
right angle are known. 

1. Given a = 42° 19' 12", e = 78° 16' 24". First draw a 
figure like Fig. 37. It is a common mistake for a beo-inner 
to glance at the two given parts in a triangle and then ask 
himself, " Which is the middle part ?" The proper method of 
procedure is as follows : Take any one of the unknown parts 
in connection with the two given parts and then ask, " Which 
of these three parts is the middle part?" To answer the 
question he must pick out that one of the trio under consider- 
ation which is either opposite to or adjacent to both the other 
parts. In this way let us consider b in connection with the 
given parts, a and e. We see from Figs. 37 and 38 that a and 
b are adjacent to each other, while e is opposite to both of 
them. Hence e must be the middle part. The first rule, there- 
fore, is the one that must be applied. And we have 

sin (co. e) = cos a cos 6, 
or cos e = cos a cos b : 

7 cos e , N 

cos b = . (a) 

cos a 

Now, in precisely the same way let A, e, and a be considered. 
We see from the figure that a is the middle part, being opposite 
to both A and e. Hence, again, by Eule I., we have 

sin a = cos (co. A) cos (co. e), 
or sin a = sin A sin e : 



144 SPHERICAL TRIGONOMETRY. 

, sin a 
sin A = — . (/?) 

sin c v J 

Finally, let B 7 c, and a be considered. Here B is the middle 
part, being adjacent to both c and a. Hence we have, by 
EulelL, 

sin (co. B) = tan a tan (co. c), 
or cos B = tan a cot c. (j) 

Thus we have found a formula for each of the unknown quanti- 
ties, b, A, and B, in terms of the two given parts, c and a. 
Special attention is called to this feature of the solution. If 
each unknown part is found only from the given ones, as can 
always be done, then no error that may be made in the deter- 
mination of one part is transmitted to another ; an advantage 
that is obvious. 

In order to check the work, we make up the formula 
between the three parts found and see whether it is satisfied 
by our results. Here our check formula is between A, B 1 
and 6, of which B is the middle part, and we have 
cos B = cos b sin A. 

The logarithmic work is as follows : 

log cos c = 9.30802 log sin a = 9.82819 

log cos a = 9.86888 log sin c = 9.99084 

by (a), log cos b = 9.43914 by (/?), log sin A = 9.83735 

6 = 74° 2 / 43" . A = ( 43° 26' 32" 

(136° 33' 28" 

log tan a = 9.95931 log cos b = 9.43914 

log cot c =9.31717 log sin A = 9.83735 

by (7) , log cos B = 9.27648 log cos B = 9.27649 check. 
J g = 79°6 / 18" 

Since A is determined by its sine, the solution gives two 
values for that part ; but I., § 108, shows that the smaller of 
these values is the correct one. Hence our results are 
b = 74° 21 43", A = 43° 26' 32", B = 79° 6 r 18". 

If the student will arrange the scheme of his work as above 
before looking up any of the logarithms, he will be able by 
two openings of the tables (as the tables are generally ar- 



SOLUTION OF SPHERICAL EIGHT TRIANGLES. 145 

ranged) to take out all six of the logarithms needed for the 
solution. 

2. Given b = 126° 38' 12", A = 67° 3' -12". 

First take c with b and A. We have, by Eule II., 

cos A = tan b cot c : 

„ . „ cos A , v 

COt C = . (a) 

tan 6 
Next take B, and, by Eule I., 

cos B = cos b sin A Q3) 

Finally take a, and, by Eule II., 

sin b = tan a cot ^. : 

, sin b f >. 

tana = — — -. 00 

cot A 

Check formula, 

cos 5 = tan a cot c. 

Logarithmic work : Whenever any of the parts of the tri- 
angle are in the second quadrant, great care must be taken to 
cany the algebraic signs of the functions with the work. 

log cos A = 9.59078 log cos b = (— ) 9.77578 

log tan a = (— ) 10.12863 log sin A = 9.96422 

by (£), log cos B = (— ) 9.74000 
B= 123° 20' 9" 

log tan a = 10.27785 
log cot c = (— ) 9.46215 
by (y), log tan a = 10.27785 log cos B = (— ) 9.74000 

a= 62° 11' 32" check. 

3. Given A = 113° 9' 12", B = 130° 18' 18". 
The formulae as given by the rules are 



cos A 
cos B 



log cot c 


= (■ 


-) 


9.46215 


c 




106 


;o 9/ 48 // 


log sin b 


■=. 




9.90441 


log cot A 


= 




9.62656 



cos c = cot A cot 5. 




O) 


cos a sin J5 : . \ cos a - 


_ cos A 
sin 5 


O) 


cos 6 sin A : .°. cos b - 


cos B 
sin J. 


(£> 


cos c = cos a cos 6. 




check. 


13 







146 



SPHERICAL TRIGONOMETRY. 



Logarithmic work : 

log cot A = (— ) 9.63109 

log cot B = (— ) 9.92850 

by (a), log cos c = 9.55959 

c = 68° 43' 54" 

log cos B = (— ) 9.81081 

log sin A = 9.96353 

by(y), log cos 6 = (— ) 9.84728 

b = 134° 42' 37" 



log cos A = (— ) 9.59461 

log sin B = 9.88231 

by (/?), log cos a = (— ) 9.71230 

a = 121° 2' 11" 

Jog cos a = (— ) 9.71230 

log cos b = (—) 9.84728 

log cos c = 9.55958 

check. 



4. Given 5 = 76° 12' 12", b = 37° 36'' 24". 

This example deserves special notice, as it is the type of a 
class having two solutions. All right triangles in which an 
angle and its opposite side are given share this peculiarity, due 
to the fact that the unknown parts are all determined by their 
sines. 

The rules give the following formulae : 

sin a = tan b cot B. (a) 

sin b ,a. 



sin b = sin c sin B : 



sin c 



cos B = cos b sin A: .-. sin A 
sin a = sin c sin A. 



Logarithmic work : 



by (a), 



log tan b — 




9.88665 


log cot B = 




9.39016 


log 


sin a = 




9.27681 


. °. 


Of! = 


10 c 


54' 12" 




a 2 = 


169 c 


5' 48" 


log 


cos i? = 




9.37745 


log 


cos b = 




9.89884 


log 


sin A = 




9.47861 




Ai = 


17 c 


SI' 10" 




A = 


162° 28 r 50" 



by (/?), 



sin i? 




\rj 


. cos 5 
cos b ' 




(r) 

check. 


log sin 6 = 




9.78550 


log sin B = 




9.98729 


log sin c = 


9.79821 


c x = 


38 c 


55' 45" 


c 2 = 


141 c 


4' 15" 


log sin c = 




9.79821 


log sin A = 




9.47861 


log sin a = 




9.27682 
check. 



by (7), 



The corresponding parts of the two results are indicated by the 
same subscript. 



SOLUTION OF SPHERICAL RIGHT TRIANGLES. 147 

110. Having shown how spherical right triangles are solved, 
it will be well to look for a moment at the limitations that exist 

between the parts. Thus from (147) we have sin c = — — . 

sin A 

Now, sin c must be less than unity. Hence sin a < sin A. A 

similar relation exists, of course, between sin b and sin B. 

Hence the oblique angles of a spherical right triangle must be 

nearer 90° thayi the sides opposite to them respectively. 

Since also sin A — ?H! — we can show in the same way that: 
sin c 

The hypothenuse of a spherical right triangle is always nearer 

90° than the sides. 

It can be proved also, as follows, that : The sum of the two 

oblique angles of any spherical right triangle is greater than 90° 

and less than 270°, and their difference is less than 90°. 

cos ( A -j- B) = cos A cos B — sin A sin B, 

cosJA + B) = cotAootB _ lt 

sin A sin B 

by (154), =cosc — 1. 

Now, sin A sin B is always -f-> hence the sign of cos (A -j- B) 
is the same as that of cos c — 1, and the latter is always — , 
since cos c is numerically less than unity. Hence A -f- B is > 
90° and < 270°. 

Again, c ^-i^-ZJl = cot A cot B + 1, 

6 ' sin A sin B ' ' 

= cos c -j- 1 ; 

whence, reasoning as before, it appears that cos (A — B) is 
always -{- ; and therefore A — B is <[ 90°. 

111. Quadrantal and Isosceles Triangles. The polar triangle 
of a quadrantal triangle is a right triangle (§ 91) ; we can 
therefore always solve a quadrantal triangle by solving its 
polar triangle. 

An isosceles triangle can always be solved by dividing it 
into two equal right triangles by means of an arc of a great 
circle perpendicular to the base at its middle point. 



148 SPHERICAL TRIGONOMETRY. 

EXAMPLES. 

Solve each of the following triangles : 

1. c = 140°, a = 20°. 

2. a = 75° 5' 18", B = 35° 29' 36". 

3. ^1 = 100°, a = 112°. 

4. a = 138° 4', 6 = 109° 41'. 

5. 6 = 40° 31' 20", c = 61° 4' 50". 

6. b = 137° 3' 39", A = 147° 2' 28". 

7. « = 59° 28', A = 66° 7'. 

8. a = 36° 12', 6 = 57° 8'. 

9. c = 90°, A == 131° 30', i? = 120° 32'. 

10. c = 90°, a = 46° 38' 46", jB = 101° 6' 22". 

In the following isosceles triangles b and c are always supposed 
to be the equal sides. Solve each one. 

11. a = 32°, B = 97° 39'. 

12. A = 55° 14', b = c = 109° 41'. 

13. a = 41° 38', A = 57° 47'. (Two solutions.) 

14. ^=72° 36', c = 49°31 / . 

15. Show that the three sides of a spherical right triangle must 
be all less than 90°, or two of them must be greater than 90°. 

16. Show that a side and the hypothenuse are in the same or 
different quadrants, according as the included angle is less than 
or greater than a right angle. 

17. If A = a, show that the remaining parts are all equal to 90°. 

18. The sides of an equilateral spherical triangle are equal to a. 
Find the formula for the angles. 

19. A ship starts due east from a port in 41° N. latitude and sails 
on an arc of a great circle of the earth 38° 37'. What is her latitude 
at the end of the journey, and in what direction is she sailing? 

20. An observer takes the altitudes of two stars, one due north 
and the other due west, and finds them to be 52° 9' and 36° 37' 
respectively. Find the angular distance between the stars. 

Prove the following relations between the parts of a spherical 
right triangle : 

2 1. sin 2 \c = sin 2 \a cos 2 \b -\- cos 2 |a sin 2 \b. 

22. tan \ (c + a) tan J (c — a) = tan 2 lb. 

23. sin (c — a) = sin b cos a tan }B. 
sin (c — a) 



SOLUTION OF SPHERICAL RIGHT TRIANGLES. 149 

24. sin 2« sin 26 = 4 cos A cos B sin 2 e. 

25. sin (c — 6) = tan 2 M sin (6 + e). 

26. sin 2 a sin 2 6 = sin 2 a -f sin 2 6 — sin 2 c. 

27. If i> is the perpendicular from C to the hypothenuse, show 

that 

(1) cos 2 p = cos 2 A + cos 2 B. 

(2) cot 2 p = cot 2 a + cot 2 b. 

(3) sin 2 p sin 2 c = sin 2 a + sin 2 b — sin 2 a 

28. sin^sini? = 2sillssill ( s ^ . 

sin 2 c 

29. sin (^L + J3) = cos a + cos 6 

1 -f- cos a cos b 

30. sin (c ~f a) = sin 6 cos a cot £i?. 
sin (c -\- a) = tan 6 cos c cot £2?. 

3 1. tan 2 (45° - U) = ten ^ c " g) . 

tan He + a) 

32. tan 2 M- Sin l C ~^ . 

sm (c -f 6) 

33. tan 2 |c = cos ^+^> 

cos {A — B) 

« 4 cos a sin 2^1 

cos b sin 2B 

35. tan S= —cot Aa cot £6 [S= } (A + B + C)]. 

36. In a quad ran tal triangle in which c = 90° sfiow that 

(1) tan a tan 6 -f sec C= 0. 

(2) cos 2 p= cosacos6 , 

COS a COS /5 

where p is the perpendicular from Cto the side c, and a and /? are 
the parts into which the perpendicular divides C. 

In the two following examples x and y represent respectively 
the internal and external bisectors of the right angle (C) of a 
spherical right triangle. Prove the relations given. 

37. 2 cot 2 x = (cot a + cot b) 2 . 2 cot 2 y = (cot a — cot b)\ 

38. 2 cos MN= (cot 2 a — cot 2 6) sin x sin y, where J/ and iVare 
the points in which the bisectors meet the hypothenuse. 

39. If 6 be the angle between the bisector of C and the perpen- 
dicular from Cto c, show that tan 4> = siu (ff ~ b \ 

sin {a + 6) 

13* 



CHAPTEE XL 

THE SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 

112. It has already been pointed out, § 87, that a spherical 
triangle can be solved when any three parts are given. We 
therefore divide the problem into six cases, as follows : 

Case I. Given two sides and the included angle. 

Case II. Given two angles and the included side. 

Case III. Given two sides and the angle opposite one of them. 

Case IV. Given two angles and the side opposite one of them. 

Case V. Given the three sides. 

Case VI. Given the three angles. 

Each of the first four cases is solved below in two ways, (1) 
by Napier's Analogies, and (2) by dividing the triangle into 
the sum or difference of two right triangles by a perpendicular 
from one of the vertices to the opposite side. Cases V. and VI. 
are each solved in only one way, by means of (134) and (135) 
for the former, and (140) and (141) for the latter. These cases 
can also be solved by dropping a perpendicular, but the direct 
method is preferable. 

It frequently happens that when one part of a spherical 
triangle is determined by the sine, we are in doubt whether to 
take it in the first or second quadrant. In such cases the fol- 
lowing principles will be of assistance in determining which 
value is the correct one. The first four are well-known 
geometric truths. 

(a) The greater side is opposite the greater angle, and con- 
versely. 

(b) Each side is less than the sum of the other two. 

(c) fl-fHK 360°. 

(d) A + B + G > 180°. 
150 



SOLUTION" OF SPHERICAL OBLIQUE TRIANGLES. 151 

(e) Each angle is greater than the difference between 180° 
and the sum of the other two angles. 
For A-\- B+ C> 180°, 

A > 180° — (£ + C). 
If, however, B -f > 180°, we have in the polar triangle to 
ABC 

a' < &' -f c' ; 

180° — A < 180° — B 4- 180°— C, 

or —.A < 1S0° — (B-\- C); 

A > (B+ C) — 180°. 

(/) A side (or angle) which differs more from 90° than 

another side (or angle) is in the same quadrant as its opposite 

angle (or side). 

The proof for the side is as follows : 

T-, /-ion\ i cos a — cos b cos c 

.From (130), cos A = — . 

sin b sin c 

The denominator of this fraction is always positive, and if a 
differs more from 90° than either b or c, we have, neglecting 
signs, either 

cos a > cos b, or > cos c ; 

cos a > cos b cos c. 

Hence the sign of the numerator and therefore that of the 

fraction is the same as that of cos a. Therefore cos A and cos 

a have the same sign, and A and a are in the same quadrant. 

The proof for the angles is derived in the same way from 

cos A + cos B cos C , -, ^ 

cos a = =!= . (see 136) 

sin B sin C 

It follows from this that only the side and opposite angle 
nearest in value to 90° can be in different quadrants. 

(g) The half sum of any two sides is in the same quadrant 
as the half sum of the opposite angles. 

This follows from the second of Gauss's equations (146) 
cos he cos 2- (A -\- B) = cos J (a -j- b) sin iC. 
For cos \c and sin \ C are always both positive. Hence cos J 
(A -\- B) and cos I (a -\- b) must always have the same sign. 



152 



SPHERICAL TRIGONOMETRY. 



113. Case I. Given Two Sides and the included 
Angle, a, b, and C. 

Solution by Napier's Analogies. 

cos J (a — b) 



tan § (A -f B) 



cos J (a -J- 6) 



cot *C, 



tan J (A — B) 



_ Tiv _ sin | (q — 6) 



sin %(a-\-b) 



cot J C, 



tan \e = sin * ^ + ^} tan J (a — b\ 
sin | (A — B) 



w 

09 



sm a 



Bin 



(7Aec#. 

sm A sin jS 

Example : 

a = 121° 17' 18", b = 

a + b = 197° 48' 36" 

a — b= 44° 40' 0" 

log cos J (a — 6) = 
colog cos i (a + 6) = (— ) 0.81024 
log cot § C = 10.32932 

log tan i(A + B) =Y~ 

%{A + B) = 
i(A-B) = 

logsinH^ + ,8) = 9.99867 

colog sin }{A — B)= 0.19697 

log tan i {a — 6) = 9.61472 
log tan \c 



sm c 



sin C 



C= 50° 12' 12' 



98° 54" 18" 
22° 23' 0" 



76° 31' 18", 
i (a + 6) 

9.96598 log sin J (a — 6) 
colog sin i (a + 6) 
log cot ^ C 
log tan J (^1 — B) 
{A 



(7=25° 6' 6" 

9.58070 

0.00527 

10.32932 



-) 11.10554 
94° 29' 3" 
39° 26' 51" 



log sin a = 9.93175 
log sin A = 9.85743 



9.81036 

Check. 
log sin 6 = 9.98787 
9.91356 
^07431 



he 
c 



9.91529 

133° 55' 54" 

55° 2' 12" 



32° 52' 13" 
65° 44' 26" 



loi 



.95985 



.07431 



.07432 

114. Solution by a perpendicular. 

In performing the solution by this method care must be 
taken to draw the perpendicular so that two of the given parts 
shall be in one of the right triangles thus formed. Denote the 
parts of the divided side by <p and <p 1 and the parts of the 
divided angle by and V taking <p and in the right triangle 
which has the two known parts. 



SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 153 

Example : Given b, c, and A. Draw the perpendicular from 
B to b. Take <p and in the right triangle that contains c 
and A. The first step is to find <p and from this triangle by 
Napier's rules. It is not necessary to compute the value of p 
(the perpendicular). For <p and we have 

cos j\ 

cos A = tan <p cot c, or tan <p = — - — - (a) 

cot c 

cos c 

cos c = cot cot A or cot = , (3) 

cot A y J 

<Pi = b — <p. (j) 

In one right triangle we have 

sin p = tan ^ cot 0, 

and in the other 

sin p = tan ^ cot X : 

.-. tan tp x cot X = tan <p cot 0, 

or 

cot 0j = tan o cot cot ^ 15 ((5) 

and B = e-\-e v (e) 

Again, in the two triangles, 

sin ^ = tan p cot A and sin <p x = tan ^> cot G. 

Hence tan p = £- = ^ : 

cot A cot G 

, n sin q>. cot JL ,„>. 

cot G = 9 — , (C) 

sin p 

and cos a = cot X cot G. (tj) 

Check. sin c? 1 = sin a sin r 

Thus, if 6 = 121° 17' 18", e = 76° 31' 18", ^1 = 50° 12' 12", we have 

log cos A = 9.80622 log cos 6 = 9.36750 

log cot c = 9.37963 log cot A = 9.92068 

by («), log tan <p = 10.42659 by (/?), log cot = 9.44682 

= 69° 28' 15" log tan <{> = 10.42659 

b = 121° 17' 18" log cot & = 9.89566 

by (y), ^ = 51° 49' 3" by (d), log cot d x = 9.76907 

6 = 74° 22' 10" log cot C — 9.84463 

X = 59° 33' 43" by (?), log cos a = 9.61370 




by CO, •# =133°55 / 53 // a = 65° 44' 26' 



154 SPHERICAL TRIGONOMETRY. 



by (Q 3 

Result 



log sin fa = 9.89545 








log cot A = 9.92068 






Check. 


colog sin <p = 0.02850 






log sin a = 9.95985 


, log cot C = 9.84463 






log sin 6 1 = 9.93560 


/ C = 55° 2' 13" 






log sin 0j = 9.89545 


t. i B =133° 55' 53" 








[ a = 65° 44' 26" 








EXAMPLES. 




1. a = 79° 6' 10", 6 = 112° 


13' 


35", 


C = 86° 14' 19". 


2. o = 121° 12' 0", 6= 76° 


37' 


0", 


(7 = 147° 23' 0". 


3. 6= 64° 23' 0", c = 100° 


49' 


0", 


J. == 95° 38' 0". 


4. a= 68° 20' 21", c = 52° 


18' 


13", 


^ = 117° 12' 52". 



115. Case H Given Two Angles and the included 

Side, A, B, and c. 

Solution by Napier's Analogies. 

tan H « + *) = :° :*g7g tanH (.) 

cot i C= B ! p *( ff + f) tan i (A - B). GO 

sin i (a — 6) 

~ 7 7 sin a sin 6 sin c 





sin A sic 


l B sin G 




Example : 








A = 78° 19', B -- 


= 36° 15', c = 


= 112° 38'. 




A — B= 42° 4' 


*(-4- 


-B) = 21° 2' 


\c = 56° 19' 


^4 + ^ = 114° 34' 


H^ + ^)=57°17 / 




log cos \ (A — B) = 


9.97005 


log sin i (A — 


-B) = 9.55499 


colog cos \ {A + B) = 


0.26722 


colog sin h{A + B) = 0.07502 


log tan Jc = 


10.17620 


log tan \c 
log tan \{a — 


= 10.17620 


log tan | (a + 6) = 


10.41347 


- b) = 9.80621 


J (a + 6) = 


68° 53' 45" 




(a = 101° 31' 0" 
(6 = 36° 16' 30" 


*(a-&) = 


32° 37' 15" 




log sin I (a + 6) = 


9.96985 






colog sin £ (a — b) = 


0.26835 






log tan £ U — B) = 


9.58493 




*C= 56° 21' 27" 


log cot ^C = 


9.82313 




C= 112° 42' 54" 



SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 155 

Check. 
log sin a == 9.99117 log sin b = 9.77207 log sin c = 9.96520 

log sin y! = 9.990 91 log sin B = 9.77181 log sin C = 9.96494 

.00026 .00026 .00026 

116. Solution by a perpendicular. 

In drawing the perpendicular we are guided by the same 
considerations as in Case I., and we letter the figure in the 
same way. 

Example : Given A, B, and c. 

See Fig. 39. The formulae for solution are derived precisely 
as in Case I., as follows : 

a 4- i cos A , x 

cos A = tan <p cot c : .-. tan <p = , (a) 

cot c 

cos c 

cos c = cot d cot a : . • . cot — -, ( t 3) 

cot A vy 

o x = B-e. ( r ) 

As before, p. 153, tan c x cot ^ = tan <p cot : 

tan <p x = tan cr cot tan 15 ((5) 

and 6 = ^-4- cr r (e) 

Now, cos = tan £) cot c, and cos X = tan p cot a : 

cos cos 0, 



tan p = - 


** — i • 

cot c cot a ' 


hence cot a - 


_ cos d x cot c ,^ 




cos e 


Finally, cos G= 


= tan <r x cot a. (jj) 


Check. cos 6 X - 


= sin C cos cr r 


Thus, if A = 78° 19', B = 36° 


15', e = 112°38', we have 


log cos A = 9.30643 


log cos c = (— ) 9.58527 


log cot c = (— ) 9.62008 


log cot A = 9.31552 


by (a), log tan = (— ) 9.68635 


by (/3), log cot 6 = (—) 10.26975 


log cot ^ = (— ) 10.26975 


6 = 151° 44' 56"* 


log tan X = 10.32152 


B= 36° 15' 0" 


by ((J), log tan ^ = 10.27762 


by(>), 0i=— H5°29 , 56 // 


logcota = (— ) 9.30912 


= 154° 5' 43" 


by (?), log cos (7= (— ) 9.58674 


0! =—117° 49' 14" 


C= 112°42 , 52 // 


by 0), 6 = 36° 16' 29" 



* Since 6 is larger than 2?, the perpendicular from B to b falls outside 
the triangle, and both d x and <p l will he negative. 



156 SPHERICAL TRIGONOMETRY. 

log cos 6 1 =(— ) 9.63396 

log cot c = (— ) 9.62008 

Check. colog cos 6 = (_) 0.05508 

log sin C= 9.96494 by (Q, log cot a = (— ) 9.30912 

log cos #! = (—) 9.66904 r a = 101° 31' 1" 

logcos^ = (— ) 9.63398 BesultAb = 36° 16' 29" 

[C = 112° 42' 52" 

EXAMPLES. 

1. A = 65° 23' 0", ^ = 101° 7' 0", c = 132° 12' 0". 

2. B = 46° 7' 0", (7= 56° 28' 0", a = 132° 46' 0". 

3. ^4 =121° 36' 18", = 34° 15' 0", 6= 50° 10' 30". 

4. A = 56° 16' 15", B — 45° 4' 45", c = 96° 20' 33". 

117. Case m. Given Two Sides and an Angle 
opposite one of them, a, b, and A. 

Solution by Napier's Analogies. 

. t> sin b sin A , . 

sin B = : , ( a ) 

sin a 

cot i 0= COsKa + 5) tan i(A + B). M 

cos J (a — 6) 

sin G sin J. 

The fact that B, the first part found, is determined by its 
sine, and will, therefore, have two values, often leads to two 
solutions to thi3 problem. The following considerations will 
show how to determine beforehand, from the data, whether to 
look for one solution or two. We have seen, § 112 (/), that a 
side which differs more from 90° than another side is in the 
same quadrant as its opposite angle. Hence 

If b differs more from 90° than a, B must be taken in the same 
quadrant as b, and there will be but one solution. 

We can also show that 

If b differs less from 90° than a, there will be two solutions. 



SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 157 



From (125) we have sin c 



cos a — cos b cos c 



sin b cos A 

Since a differs more from 90° than b, we have, neglecting 
signs, cos a > cos b and hence cos a > cos b cos c. Hence 
the sign of the numerator is the same as the sign of cos a ; 
while, sin b being positive, the sign of the denominator is the 
same as that of cos A. Now, cos a and cos A have the same 
sign, since a and A are in the same quadrant, § 112 (/). Hence 
under the conditions assumed sin c is positive, and there will 
be two values of c, both less than 180°. AVe have, therefore, 
two values for both B and c, and hence also for C. 

The solution will be impossible if a and A are given in 
different quadrants and a differs more from 90° than b. § 112 
(/) ; or if sin b sin A > sin a, for in that case sin B > 1. 

(For a very complete discussion of this question see McClel- 
land and Preston's Spherical Trigonometry, Part I. pp. 137- 
142.) 

Example : 

a = 67° 35' 12", b = 58° 36' 6", A = 101° 17' 48". As b differs more 
from 90° than a, there will be but one solution, and the value of B 
must be less than 90°. 



I {a + b) = 63° 5' 39" 
i{a — b) = 4° 29' 33" 



log sin b = 


9.93124 


log sin A = 


9.99150 


colog sin a = 


0.03411 


log sin B = 


9.95685 


B =64 


3 52' 50" 


log tan % (a + b) = 


10.29460 


log cos ${A + B) = 


9.08039 


colog cos i (A — B) = 


0.02231 


log tan Jc = 


9.39730 


ie = 


14° 0' 59" 


c = 


28° V 58" 


flog sin c = 


9.67207 


Check. J log sin C= 


9.69769 



S3 C 



19 / 



£ (A — B) = 18° 12' 29" 

log tan J (^L + 5) = 10.91645 

log cos J (a + 6) = 9.65564 

colog cos i (a — 6) = 0.00134 

log cot £C = 10.57343 

|C=14°57' 5" 

C=29°54'10" 

log sin a = 9.96589 

log sin J. = 9.99150 



{ 9.97438 

118. Solution by a perpendicular. 
Given a, b, and A 

14 



9.97439 



158 



SPHERICAL TRIGONOMETRY. 



Fig. 40. 



The two figures represent the way in which it is sometimes 
possible to construct two triangles in this case with the same 
three given parts. The values of <p, <p v 6, and X in the two 
figures are of course the same, and the two triangles BPC 
have their several parts respectively equal. But in the one 
case c = tp -f- <p x and 
C = + V and in the 
other c = tp — <p and 
G = 6 — V Hence in 
solving such a triangle 
we must always look 
out for a double solu- 
tion ; but if <p — tp x and 
— 6 X are negative, or if tp -f- <p x and -}- J are > 180°, there 
is but one solution. If both of these conditions are fulfilled, 
there is no solution. The formulse, derived by means of 
Napier's Eules, are as follows : 

cos A = tan tp cot 




cos b = cot 6 cot A 



, cos^l 

tan tp = , 

cot b 

cos b 



cot = 



cos a = cos tp x cos p, cos 



cot A 
COS tp cos p 



(«) 

(JO 



cos p 



cos a 



cos b 



or 



COS tp x 



COS tp x COS tp' 

cos a cos tp 



cos 6 



cos b 
cot b tan p, cos 6 X = 



(r) 



cot a tan p : 



tanp 



cos 6 



cos t 



or 



cot b cot <2 
cos cot a 



cos #, = 



cot b 



Check. 



c = <p ±<p 1 ; 

C=0±0 X . 
cos B = cot a tan p r 
cos = sin B cos ^ r 



« 



SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 159 

Example : a = 67° 35' 12", b = 58° 36' 6", A 

log cos A ={—) 9.29201 
9.78559 



log tan </> = (— ) 9.50642 

log cos a = 9.58125 

log cos</> =(— ) 9.97872 

colog cos b = 0.28317 



log cos X =(— ) 9.84314 

<$> = 162° 12' 24" 
<p x = 134° 10' 28" 



c 1 = 296° 22' 52" 
c,= 28° 1'56" 



56' 6", A = 


101° 17' 


48". 


log cos b 


= 


9.71683 


log cot A 


= (-) 


9.30051 


log cot 6 


= (-) 


10.41632 


log cot a 


= 


9.61537 


log cos 


= (-) 


9.97022 


colog cot b 


= 


0.21441 


log COS 0j 


= (-) 


9.80000 





= 159° 


1'19" 


0i 


= 129° 


7' 15" 


Ci 


= 288° 


8' 34" 


C 2 


= 29° 


54' 4" 



But the values of c x and C x are impossible. Hence there is but 
one solution, and 

c = 28° V 56", C= 29° 54' 4". 

In applying (/?) and the check formula the negative value of <p x 
must be used, since it is that that gives the only possible value 
of c. 





Check. 


log cot a 


= 9.61537 log sin ^ = 9.95685 


log tan (- 


-^) = 10.01251 log cos (— ^) = (— ) 9.84314 


log cos B 


= 9.62788 log cos 2 =(—) 9.79999 




B = 64° 52' 51" 




EXAMPLES. 


1. a = 


99° 40' 0", b =64° 28' 0", ^4 = 95° 38' 0". 


2. b = 


98° 17' 0", e = 74° 37' 0", C = 61° 13' 0". 


3. a = 


120° 30' 30", c = 69° 34' 50", A = 80° 10' 20". 


4. c = 


89° 28' 15", b =59° 50' 7", (7 = 56° 28' 28". 



119. Case IV. Given Two Angles and the Side 
opposite one of them, A, B, and a. 

A triangle given in this way may be reduced to the previous 
case by taking its polar triangle. Hence we meet here with 
the same peculiarities that we found in Case III., so far as 
single and double solutions of the triangle are concerned. 
The principles set forth in § 117, adapted to this case, will 
read as follows : 



160 



SPHERICAL TRIGONOMETRY. 



If B differs more from 90° than A, b will be in the same quad- 
rant as B, and there will be but one solution. 

If B differs less from 90° than A, there will be two solutions. 
Let the student prove this by means of the formula 

. n cos A -j- cos B cos C 

sin u — — — . 

sin B cos a 

If A and a are given in different quadrants, and A differs more 
from 90° than B, or if sin B sin a > sin A, the solution is 
impossible. 

Solution by Napier's Analogies. 

sin B sin a 



sin b 



sin A 



tan ie = cos » ^ + -8) tan i (a + V), 

cos i (A — £) y ^ J ' 

cot iC= - C08 H a + b ) tan i (A + B). 
cos i (a — b) 



0) 

GO 



Check. 



sin a 



sin c 



sin sin A 



Example : A = 139° 54' 36", ^ = 61° 37' 54", a = 150° 17' 24". 
Since B differs less from 90° than A, there will be two solutions 
to this example. 



log sin B 
log sin a 
colog sin A 
log sin b 

b, 



9.94444 
9.69514 
0.19112 

" 9.83070 

42° 37' 23" 

137° 22' 37" 



| {A + B) = 

±{A-B) = 
I (a + b x ) = 
i (a _ 6i) = 
i(a + b 2 ) = 
* (a — 6-) = 



100° 46' 15" 
39° 8' 21" 
96° 27' 24" 
53° 50' 0" 

143° 50' 0" 
6° 27' 24" 



log cos $(A + B) = (— ) 9.27156 log cos § (a + 6 a ) = (— ) 9.05097 

colog cos J(^ — B)= 0.11035 colog cos £(« — b i) = 0.22905 

log tan ^ (a + 6 X ) = (— ) 10.94627 log tan $(A + B) = (— ) 10.72071 

logtanfo = 10.32818 log cot I C\ = 10.00073 

logcos^ + g) =(— ) 9.27156 log cos $ (a + b 2 ) = (— ) 9.90704 

colog cos |(^ — ^)= 0.11035 colog cos $ (a — b 2 ) = 0.00276 

log tan i \a + 6 a ) =(— ) 9.86392 log tan j (A + -g) = (— ) 10.72071 



log tan \c 2 



= 9.24583 logcotJC, 



10.63051 



SOLUTION OF SPHERICAL OBLIQUE Till ANGLES. 161 





hc x = 


64° 50' 25" 




|Ci= 44°57 / 7" 




2^2 = 


9° 59' 20" 




£C 2 = 13°10 / 42" 




<h = 


129° 40' 50" 




Ci= 89° 54' 14" 




c 2 = 


19° 58 / 40" 

Check 




C 2 = 26°21 / 24" 


>g sin c t ■ 


= 9.88628 


log sin c 2 = 


9.53359 


log sin a = 9.69514 


>g sin C x 


= 0.00000 


log sin C 2 = 


9.64734 


log sin A = 9.80888 



9.88628 9.88625 9.88626 

120. Solution by a perpendicular. 
Given A, B, and a. 

A figure similar to Fig. 39 or the first triangle in Fig. 40 
may be used to derive the formula?, which are as follows : 

cos B = tan <p cot a : . • . tan <p = — , (a) 

cot a 

cos a = cot 6 cot B : . • . cot = -. (/3) 

cot B 

sin (p = tan p cot 5, sin <p x = tan ^> cot A : 

sin £> sin p T 



tan j9 == 



cot B cot A 



sin £> cot A , v 

or sin ^ = £— — . O) 

cot I? 

cos B = sin cos p, cos A = sin X cos j? : 

cos B cos J. 



sin u cos ^l z >N 

or sin d x = — . (3) 



Check. 



uw p 




sin sin #/ 


sin 


*i = 


sin cos A 


eos B 




<? 


= <P + <P V 




(7 


= o+o v 


cos b : 


= cot X cot J.. 


sin 


SV 


= sin b sin 1§ 



0) 
(0 



*0 X and X are both determined from the sine, so that each of them will 
have two values less than 180°. These values will he positive or negative 
according as sin </> x and sin 6 1 are positive or negative. Both values of each 
of these quantities must be used in (e) and (£). 
I 14* 



162 



SPHERICAL TRIGONOMETRY. 



JExa?npl 


g/ - 


,4 = 


= 139° 54 / 36", .B = 


= 61 


37' 54", a 


= 150° 17' 24". 


log cos B 


= 




9.67682 




log cos a 


= (-) 


9.93879 


log cot a 


= 




(— ) 10.24365 




log cot B 




9.73238 


log tan <f> 


= 




(— ) 9.43317 




log cot 6 


= M 


10.20641 


log sin <p 


= 




9.41777 




log sin 


= 


9.72263 


log cot A 


= 




(— ) 10.07480 




log cos J. 


= (-) 


9.88368 


colog cot 1 


3 = 




0.26762 




colog cos B = 

log sin X = ( — ) 


0.32318 


log sin <p ± 


(— ) 9.76019 


9.92949 


4> 


= 




164° 49' 49" 




6 


= 148 c 


7' 49" 


*i 


= 


{- 


—35° 8' 53" 
-144° 51' 7" 




*i 


_ f — 58 c 
t— 121 c 


13' 37" 

46' 23" 


c i 


= 




129° 40' 56" 




Ci 


= 89 c 


' 54' 12" 


c 2 


= 




19° 58' 42" 




c 2 


= 26° 21' 26" 














Check. 




log cot 6 1 


= 




(=F) 9.79196 




log sin 6 


= 


9.83071 


log cot ^4 






(— ) 10.07480 




log sin X 


= (-) 


9.92949 


log cos b 






(±) 9.86676 




log sin fa 


= (-) 


9.76020 


&i 






42° 37' 30" 










& 2 






137° 22' 30" 
















EXAMPLES. 








1. 


.4 


= 76° 6', B = 


: 85 


3 22', b = 


93° 18'. 






2. 


^L 


= 132° 16', C = 


139 


3 44', c = 


127° 30'. 






3. 


.4 


= 70° 0', B = 


:120 


3 0', b = 


100° 0'. 






4. 


.4 


= 128° 19', B = 


70 


3 0', a = 


142° 16'. 





121. Case V. Given the Three Sides, a, b, and c. 
This case is most conveniently solved by (134) and (135), 
which are repeated here. 

t^ r _ sin (s — a) sin (s — b) sin (s — c) 



Check. 





sin 


s 


fnn n 


tan 


r 




sin (s - 


- a y 


tan hB -. 


tan 
sin (s - 


r 

- b y 


+*\n I C 


tan 


r 




sin (s - 


-«)" 


sin a 


sin b 


sin c 



sin A sinB sin C 



SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 163 



Example : a = 112° 6' 42", b = 127° 39' 12", c = 71° 12 / 42". 
2s = 310° 58' 36" colog sin s = 0.38208 

s = 155° 29' 18" log sin (s — a)= 9.83682 

s — a = 43° 22 / 36" log sin (s — 6) = 9.66924 

s _ b = 27° 50' 6" log sin (s — c) = 9.99783 

s _ c = 84° 16' 36" log tan 2 r = 9.88597 



log tan r = 9.94298 

log sin (« — a) = 9.83682 

logtan^ = 10.10616 
%4 = 51° 56' 2" 
.4 = 103 J 52' 4" 



log sin a = 9.96683 
log sin A = 9.98715 



log tan r = 9.94298 

log sin (* — &) = 9.66924 

log tan 3^5 = 10.27374 

Y 2 B= 61° 58' 4" 

B = 123° 56' 8" 

Check. 

log sin b = 9.89857 
log sin B = 9.91890 



log tan r = 9.94298 

log sin (s — C ) = 9.99783 
log tan y 2 C = 9.94515 
3^C=41°23'29" 
C= 82° 46' 58" 



log sin c : 
log sin C 



9.97968 



9.97967 



9.97622 
9.99655 
9.97967 



1. a = 106° 48' 6' 

2. a = 123° 19' 0' 



EXAMPLES. 

6 = 85° 4' 30", e 
b = 67° 38' 0" e 



97° 55' 0". 
164° 12' 0". 



122. Case VI. Given the Three Angles, A, B, and a 
Here we use (140) and (141). 

— cos S 



tan 2 R 



Check. 



cos (£ — A) cos (S—B) cos (£— C)' 
tan £<2 = tan R cos (# — A), 
tan *6 = tan R cos (& — 5), 
tan lc = tan jR cos (£ — 0). 
sin a sin 6 sin c 



sin A sin i? sin C' 
The logarithmic work may be arranged as in Case Y. 



EXAMPLES. 

87° 13', C= 96° 48'. Find a, 5, and c. 



1. ^4 = 106° 19', 5 

2. ^1 = 64° 19', ^ = 112° 27', C= 100° 00'. Find a, 6, and c. 

3. One point on the earth's surface is in latitude N. 38° 47', lon- 



54° 6'. What is their distance apart in geographical miles ? 



164 SPHERICAL TRIGONOMETRY. 

4. Four lines, OA, OB, OC, and OD, meet at the point O and 
make the following angles : AOB = 47° 12', BOC=72° 16', CO A = 
68° 14', AOB = 36° 19', and BOD = 49° 21 / . Find the angle COD. 

5. a = 106° 48' 6", 6 = 85° 4' 30", c = 97° 55'. Find the three 
perpendiculars from the vertices to the opposite sides.* 

6. Given a = 42°, B = 67°, 6 + e = 112° : solve. 

7. Given a = 38°, ^ = 101°, A+ C= 136° : solve. 

8. If Cj and c 2 be the two values of the third side, when A, «, 
and 5 are given and there are two solutions, show that 

tan I e ± tan Jc 2 = tan J (b — a) tan J (6 -f «)• 

* Use § 106, Ex. 2. There will be two results for each perpendicular, 
since from any point in a sphere it is possible to draw two arcs, supplements 
to each other, perpendicular to any great circle. 



CHAPTEE XII. 

AREA, CIRCUMSCRIBED AND INSCRIBED CIRCLES. 

123. Area of a Spherical Triangle. It is shown in 
geometry that the area of a spherical triangle is equal to its 
spherical excess. That is, if the amount by which the sum of 
the angles of a triangle exceeds 180° be expressed in terms of 
a right angle as a unit, this same expression will be the area 
of the triangle in terms of the trirectangular triangle as a 
unit. 

Thus, if 

A = 60°, 5 = 80°, C=110°, 

(A + B -f C) — 180° = 70° = I of a right angle : 

area (JT) = % of a trirectangular triangle. 

Now, since the trirectangular triangle is i of the surface of 
the sphere, its area is l~r 2 . On any sphere whose radius is r 
feet, therefore, the area of the above triangle will be 
K=z -fg-r 2 square feet. 
If we use the notation S = i ( A -f- B -f- C), we have 

K=2S — 7T J when r = 1, (157) 

or, more generally, K= r 2 (2£— -), (158) 

where S is expressed in radians. 

Example : Upon a sphere whose radius = 10 feet a triangle has 
A = 72° 32', B = 58° 6', C= 101° 9'. Find K. 

13907_ 



2S= 231° 47' = 1390: 



10800 



2^-^ = 13907^ „_3107 



10800 10800 

By (158), 

log 3107 -=3.49234 
K= 100 X ^-i 7T square feet. colog 108 = 7.96658 

10800 -, ,,._.,_ 

log rr = .49715 



K~ 90.38 square feet. log K = 1.95607 

165 



166 SPHERICAL TRIGONOMETRY. 

124. De Gua's and Lhuillier's Formulae. 

If r = unity, the above expression for K becomes K= 2S — n. 
Hence iK= S — |tt, and we have 

sin iK= — cos S, cos %K= sin S, tan %K= — cot S. (159) 
From (137) and (138) we have, after reduction, 

sin %a sin §j — cos S. 

cos \c sin C 

-cosS= aiuiasinib mnO. 

cos \a 

Now, sin C= 2 sin §(7 cos IG 



2y sin s sin (s — a) sin (s — 6) sin (5 — c) 
sin a sin b 

by (131) and (132) 

see § 106, Exercise 1 



2 sin \a cos la sin £6 cos £6' 



—cos S = - = sin IK. (160) 

2 cos \a cos J6 cos \c 

To find the value of sin S, we have first 

cos (S — G) = cos S cos (7 -j- sin S sin (7; 

. a cos (iS> — (7) — cos S cos (7 

sin £ = * l — . 

sin G 

Now, by (138) we have, after reduction, 

cos ja cos \b cos (aS" — C) 

cos \c sin C 

and we have above a value for — in terms of the sides. 

sin C 

Substituting these values in the above expression for sin S, it 
reduces to 

. Q cos \a cos \b -f- sin \a sin \b cos C 

sin o — • 

cos \c 

To reduce this further we substitute the value of cos G derived 

from (125), 

n cos c — cos a cos b cos c — cos a cos b 

cos C = 



sin a sin b 4 sin \a cos \a sin |6 cos \b 



AREA, CIRCUMSCRIBED AND INSCRIBED CIRCLES. 167 

1 „™ 1 i, i cos c — cos a cos b . 
cos \a cos \b -j 

m , . . a 4 cos ha cos lb 

This gives sin & = , 

& cos \c 

_ 4 cos 2 ja cos 2 \b -f- cos e — cos a cos & 
4 cos \a cos ^ cos \c 
Finally, since 

4 cos 2 ha cos 2 £6 = (1 + cos a) (1 -f cos 6), by (58) 
we have, after collecting terms, 

sin S = 1 + C0aa + C08i + C08c = cos IK. (161) 
4 cos \a cos \b cos £c 

mi i Z^ Bill li^ 2ft 

Then tan }A 



cos \K 1 -f cos a -j- cos b -j- cos c 

This is known as De Gua's formula. 

To derive Lhuillier's formula, Ave proceed as follows: 

r> «nv * ,17 /l COS ^JT 1 COS ^ 

J v J \1 -f cos \ K sin \R 

but (159) cos %K= sin fif, and sin IK= — cos S; 

1 — sin S 



tan }JT: 



— cos S 



If in (161) we substitute for cos a the value 2 cos 2 ha — 1, 

derived from (58), and similar expressions for cos b and cos c, 

we have 

. a cos 2 \a 4- cos 2 £& + cos 2 lc — 1 

sm S = o i Ti i * 

2 cos -o-tf cos hb cos |c 

and 

1 . a 2 cos ha cos hb cos he — cos 2 ha — cosH?> — cos 2 *c-f-l 

I — sm o = ^— J — . 

2 cos \a cos hb cos \c 

Xow, from § 106. Exercise 1. we see that the numerator of this 
fraction is equal to 

4 sin |s sin h (s — a) sin h (s — b) sin § (s — c). 

Hence we have 

1 . „ 2 sin hs sin h (s — a~) sin h (? — V) sin h (s — c) 

I — sm £= — . 

cos ha cos ho cos he 



168 



SPHERICAL TRIGONOMETRY. 



Substituting this value in the numerator and the value of — 
cos S from (160) in the denominator of the expression for tan 
IK, we have, after reduction, 

b) sin j (s — c) 
n 



tan ! K 4 sin Is sin | (s — a) sin \ {t 

n 

_ 4 sin Is sin \ (s — a) sin \ (s — b) sin j (s — c) 



Now, 



l/sin s sin (s — a) sin (s — 6) sin (s — c) 
sin 55 sin is 



V sin s y 2 sin Js cos %s 



—77, l/tan £s, etc. 
1/2 



.-. tan £if=i/ tan Jstan^ (s — a)tan£ (s — b) tan f (s — c), 

(162) 
which is known as Lhuillier's formula. 



125. Circumscribed Circle. 

Let be the pole of the small 
circle circumscribed about the tri- 
angle ABC Draw the arcs OA, 
OB, and OC, and draw OP from 
perpendicular to BG. The tri- 
angles OBC, OCA, and OAB are 
isosceles, and BP=%a. Let It 
represent the radius OA, OB, or 
OC In the right triangle OBP 
we have, by § 107, II., 

cos OBP = cot B tan \a, 
tan \a 



Fig. 41. 




or 



Now, 

and 

Adding, 



tan .B = 

cos OBP 

OBP = B — ABO=B — BAO, 

OBP= OCP= C—ACO= C — OAC. 

20BP = B-\- C—(BAO+ OAC), 

= B+ C—A, 

= 2(S — A): 

OBP=S—A. 



AREA, CIRCUMSCRIBED AND INSCRIBED CIRCLES. 169 



Hence 



tan R = 



tan la 



tan R = 



tan R 



cos (S — Ay 

tan lb 



cos (S—B) 
tan %c 



(163) 



cos (£— C) 

If the value of tan \a, from (139), be substituted in the first of 
these expressions, we have 
1 



tan R = 



cos (S — A) 



j 



or tan R 



-v 



cos S cos (ff — A) 
cos (S—B) cos (£— C)' 

(164) 



cos £ 



0) 



Fig. 42. 



cos (S—A) cos (S—B) cos (&■ 

126. Inscribed Circle. 

Let be the pole of the circle inscribed in the triangle 
ABC. Draw the arcs ON, OM, and OQ perpendicular to 
AB, BG, and CA, at the points where they 
are tangent to the inscribed circle; and 
draw OA, OB, OC. The right triangles 
AON and AOQ have their corresponding 
parts mutually equal, since they have the 
side A in common and NO equal to OQ. 
Hence AN= A Q. Similarly BN= BM, 
and CM = CQ. We have also OA, OB, 
00 bisecting the angles A, B, C respec- 
tively. Eepresent the radius ON by r. 
Then, in the right triangle AON, 

sin AN= tan r cot \A. 
Now, AN= c — BN= c — BM, 

and AQ = b—QC=b—CM. 

Adding these equations, we have, since AN -\~ AQ = 2AN, 
2AN= b + c — {BM + CM), 
= b -f c — a = 2 (s — a); 
AN=s — a. 
Hence, from (a), sin (s — a) = tan r cot I A. 

H 15 




00 



170 SPHERICAL TRIGONOMETRY. 

We may therefore write 

tan r = sin (s — a) tan %A, } 

tan r = sin (s — b) tan %B, J- (165) 

tan r = sin (s — c) tan J C. J 
If in the first equation in (165) the value of tan \A in (133) 
be substituted, we have 



, sin (s — b) sin (s — c) 

tan r = sm (s — a) \ i 1—, — * S 

J \ sm s sin (s — a) 

or tan r = / "n (s- a) sin (s- ft) sin («-^ (16fi) 

\ sin 5 

EXAMPLES. 

1. Find the area, the radius of the circumscribed circle, and the 
radius of the inscribed circle in the two triangles in \ 121. 

2. Show that 

N 



(1) tan R = 



(2) tan r 



cos (S— A) cos (S— B) cos {S— C) J 



n 



3. Show that 

(1) tan R tan r = - cos ^ slna = - 9^A^kA j e tc. 

sin s sin A sin s sin B 

(2) tan B = 4 tan r -cos £ sins 



sin a sin b sin c sin vl sin B sin C 

4. Prove in any right triangle 

sin \K= ^iasmjb 
cos Jc 

COB^= COB * qOOB * 5 . 

cos Jc 

5. The area of an equilateral triangle is one-fourth the area of 
the sphere. What are its sides and angles? 

6. In any equilateral triangle 

tan R = 2 tan r. 

7. If ABC be an equilateral spherical triangle, P the pole of the 
circumscribed circle, and Q any point on the sphere, show that 

cos QA + cos QB + cos QC= 3 cos i? cos FQ. 



APPENDIX. 



The following list of formulae comprises most of those that are 
used constantly in making trigonometric reductions. They are 
placed together here for convenience of reference when the student 
has passed on to higher branches of mathematics. The formulae 
are numbered as in the text. 



tan cot 6 = 1. 

cos sec 0=1. 

sin esc 0=1. 
sin 2 0+ cos 2 = 1. 
1 + tan 2 = sec 2 0. 
1 -f cot 2 = esc 2 0. 

tan0 = ^, cot 6 = ™*° 



cos 0' 

sin =cos (J7r — (/), 
tan = cot Qtt — 0). 
sec =csc (in — 0), 



sin 

cos = sin (in- — 
cot = tan (Jtt — 
esc = sec ($tt — 



0). 



sin (iir + 0) = cos 0, 
sin (n — 6) = sin 0, 
sin (tt + 0) =— sin 0, 
sin (fir — 0) = —cos 0, 
sin (fir + 0) = —cos 0, 
sin (277 — 0) = —sin 0, 
sin (— 0) = —sin 0, 



cos (£tt + 0) = —sin 0, 
cos (tt— 0) = —cos 0, 

COS (tt + 6) = —COS 0, 

cos (fir— 0) = — sin 0, 
cos (fir + 0) = sin 0, 
cos (2ir — 0) = cos 0, 
cos (—0) = cos 0, 



tan ($tt + 0) 

tan (tt — 0) 
tan (tt + 0) 
tan (fir — 0) ■■ 
tan (fir + 0) ■■ 
tan (2rr — 0) 
tan (— 0) 



-cot 
-tan 
tan 
cot 0. 
-cot 0. 
-tan 
-tan 0. 



(13) 
(14) 
(15) 
(20) 

(21) 
(22) 

(23) 

(16) 
(17) 

(18) 

(39) 
(42) 
(40) 
(43) 
(41) 
(44) 
(45) 



sin (x ±y) = sin x cos y ± cos a? sin ?/. 
cos (a; ± y) = cos a; cos y =f sin as sin y. 
tan # =h tan y 



tan (a; =b #) 



1 =F tan x tan y 
! sin a; cos x. 



sin za; 
cos 2aj = cos 2 x — sin 2 x. 
2 tan x 



tan 2x 



tan 2 a; 



(48), (49) 
(50), (51) 

(52), (53) 

(54) 
(55) 

(56) 

171 



172 APPENDIX. 

sin Zx = 3 sin x — 4 sin 3 x. (67) 

cos Sx = 4 cos 3 x — 3 cos x. (68) 

1 — cos a; =2 sin 2 $x, 1 + cos x = 2 cos 2 \x. (57), (58) 

tan 2 %x = }~ Goax . (59) 

1 -f- COS X 

1 — sin x = 2 sin 2 ( Jtt — J») = 2 cos 2 (£*■ + %x). (65) 

1 + sin x = 2 sin 2 (£tt -f %x) = 2 cos 2 (£tt — Ja). (66) 

tan 2 (Itt — Js) = cot 2 (^ + Ja;) = 1 ~ siu x . 

1-fsinx 

sin (a -f- ft) tan a -f tan ft /wqx 

sin (a — j3) tan a — tan ft' 



(71) 



cos (a -f ft) 1 — tan a tan ft 

cos (a — ft) 1 -f- tan a tan ft' 

sin a + sin ft = 2 sin H a + ft) cos J (a — ft). (60) 

sin a — sin ft = 2 cos J (a + ft) sin J (a — ft). (61) 

cos a + cos ft = 2 cos I {a + ft) cos H a — ft)- ( 62 ) 

cos ft — cos a = 2 sin J (a + ft) sin I {a — ft). (63) 



ANSWERS TO EXAMPLES. 



Page 14. 

5. (1) .2326, (2) 1.1746, (3) 3.0219, (1) 5.1700, (5) 6.6020, (6) 
10.0683. 6. (1) 1.3956, (2) 9.1971. 7. 10° 6'.1, 35° 45'.4, 93° 54', 
157° 20'.4, 222° 17', 300° 17 / .5. 8. 222° 20', 64° 36'.7. 9. .0174533, 
.00029089, .000004848. 10. 613° 52'. 

Page 26. 
30. 1.2232, 2.1953. 31. 26° 35' 47". 32. 4381.3. 

Page 35. 

10. 0, 7T, 27T, 37T, . . . . n~. 11. £7T, f 7T, fTT, . . . . ^^ «T, aild 7T, 3tT, 

5tt, . . . . (2/1 + 1) it. 12. 0, 7T, 2tt, 3-, . . . . wk. 15. 5, 8, —5, f. 

Page 42. 

1. —sin 17°, —cos 59°, —tan 71°, sec 48°, etc. 

Page 43. 

2. —sin 17°, —sin 31°, —cot 19°, esc 42°, etc. 3. (1) (a — 6) sin 0, (2) 
tt(l-tan'fl-6 S ec'0 (5) tan a? (6) (7) gin ^ 

tan </» 

Page 44. 

28. 24° 5F 15" and 204° 51' 15". 29. 27° 13' 26" and 152° 46 / 34". 
30. 107° 54' 40" and 287° 54 / 40". 31. 0°, 120°, 240°. 32. 45°, 135°, 

225°, 315°. 34. ^ = 50°, c = , a = 7 sm 50 ° . 

sin 50° + cos 50° sin 50° + cos 50° 

35. ^1 = 67*°, a = , 5 = 8cos67 ^° , c = . 36. 

2 ' sin m° sin 2 67*°' sin 2 67*° 



9 



2 sin 54 c 



l/ sin 2 - 54° — cos 2 54°' ° V sin 2 54° — cos 2 54° ' 
A = 75°, a = 15 sin 75°, b = 15 cos 75°. 

Page 45. 

52. ± | ^5, ± A /5, ± A /39, ± (2 ± f) ^2. 

15* 173 



174 APPENDIX. 

Page 56. 

1. sin 15° = cos 75° = \ (/6 — /2), cos 15° = sin 75° = \ (/6 
+ /2), tan 15° = cot 75° = 2 — /3, cot 15° = tan 75° = 2 + /3. 2. 
sin 22 \° = $i/2 — /2, cos 22£° = Jl/2 + /2, tan 22 J° = /2 — 1, cot 
22J° = /2 + l. 

Page 57. 



33. sin 20 = f/2, cos 20 = £ ; sin f0 = /a — i/2, cos £0 = 
^-2- + i/2. 34. sin 20 = 1/7, cos 20 = — i; sin 30 = £/2, cos £0 
= i/14. 35. - T V 

Page 58. 

39. 0, Itt , f tt, tt, |tt, -V-tt. 40. 0, in, fvr, tt, |tt, |tt. 41. tan" 1 ± 
VI ± f/3. 42. 0, tt. 43. Itt, Itt, |tt, |tt, fvr, |tt. 45. (1) I 

(•7 + /6), (2) 1/7 + A, (3) ~^3' (4) 12 ~ 5 V2 i > (5) mn + 



a — 2 



l/(l-m 2 )(l-n 2 ), (6) v — -|. 46. (l)±J(i/5-l), (2)^/2, (3) 

1/2, (4) fc (5) 1/6, (6) -|f, (7) ma ^^ wa > (8) /*, (9) 

5/3 + V14 

18 

Page 68. 

26. (1) .73354, (2) 1.21992, (3) 1.77652. 27. (1) 2, (2) 28.784. 28. 
(1) 44°, (2) 124°. 29. 67°. 30. (1) 1.4142 and .7071, (2) 1.1756 and 
.8090, (3) 1 and .8660, (4) .8678 and .9010, (5) .7654 and .9239, (6) .6840 
and .9397, (7) .6180 and .9511, (8) .5634 and .9595. 31. 2 sin- 1 i 
= 38° 56' 31". 32. 2.58 and 397.43. 33. 28° 36' or 118° 9 / 48". 34. 
(1) 10° or 36°, (2) 1034.1 or 3962.7. 35. 3.448. 36. 20.900 and 21.636. 
37. 10 metres. 

Page 69. 

38. 4.1501 square centimetres. 39. 40 and 22 metres. 40. 83 
feet. 41. 116.06 and 196.84 feet. 42. 32° and 148°. 43. 42°. 44. 
148 feet. 45. 58°. 46. /2 : 1. 47. The sine of the larger part 
equals twice the sine of the smaller part. 48. 41° and 139°. 49. 
26° 30', 63° 30', 116° 30', and 153° 30'. 50. 50°. 51. A = 53° 7' 49", 
B = 36° 52' 11". 

Page 70. 

52. — 5- — . 53. 717.6 feet. 54. 1486 feet. 56. h = 610.67 feet, 
2 sin A 

MA = 693.89 feet. 57. 4 feet. 58. 1301.7 metres. 59. 43° 35' 5". 

60. 6557 and 1639 feet. 



APPENDIX. 



175 



Page 71. 

61. 3 metres. 62. (1) 135.85 feet, (2) 492.73 feet, (3) 361.41 feet. 
63. 40 and 90 feet. 64. 50° 19', 452.7 feet, 507.4 feet, 390.5 feet. 
65. (1) 349.29 and 608.97, i2, 461.2, (3) 409.09. The last. 66. 52.93 
miles. 

Page 92. 

1. I 63, (1) 119342, (2) 32916, (3) 1513.6, (4) 122671, (5) 934.18, (6) 
164377; g 64, (1) 19.012, [2 832.7, 3 20244800, 4) 109S.4. 2. Sides 
= 105. SI and 434.16. diagonal = 386,5, angles = 13 3 14' 35" and 109° 
57' 25". 3. Parallel sides = 174.55 and 277.09, other sides = 89.66. 
4. 3S4.17 and 341.38. 5. sin 24° : sin 36°. 6. 29. 7. Sides = 98.955 
and 73.386, angles = 70° 54' 54" and 109° 5' 6". 8. Sides = 130.56 
and 219.46, angles = 65 D 4' 8" and 114° 18' 52". 9. 328.15 feet. 



10. 607.67 feet. 11. 

tion at A = 17° 19' S". 



Page 93. 
AC= S39.12 feet, height 



12. 36.06 feet. 13. 



261.66 feet, eleva- 
Length = 116.07 feet, 



distance = 164.58 feet. 14. Length = d sirL J 9 a) sin , distance 



d sin (o — a) cos 6 

sin (0 — o) cos a 
14", 76° 49' 56". 17. 
1231.9 feet. 



sin {6 — o) cos a 
15. 15.004, or 52.488. 16. 41° 40' 4S", 61° 29 / 
135.8. 18. 1151 feet. 19. 489.24 feet. 20. 



Page 94. 
21. 1970.2 feet. 23. a =195.74, 6 = 213.75. c = 263,57. 24. A = 
41° 4S' 2", B = 100° 40' 22", c = 195.75. 25. .4 = 49° 28 / oS", 6 
= 244.4S, c = 221.69. 26. c = 30.06, 0= 62° 14'. 27. C= 129° 3' 12", 
= 110.05,6 = 67.35. 28. a =382.99, b = 349.01, c = 439.42. 29. a 
= 371.72,6 = 155.78. 30. a =10.259, b =13.341, c = 11.319. 31. a 
= 967.96, 6 = 456.7S, c = 1013.23. 32. a =256.81, 6 = 170.84, c = 
213.55. 33. B = llQ°o0', C=15°51 / , « = 562.31, 6 = 682.57, c = 
20S.92. 34. a = 156.62, 6 = 106.3S, A = 104° V 40", B = 41 c 13' 40". 
35. 6 = 227.05, c = 51.17, B = 109° 13' 51", C= 12° 17' 9". 36. a = 
34.855, 6 = 33.438. 37. a = 99,54, 6 = 161.79, c = 149.91. 38. a = 
166,50, 6 = 1S2.79, c = S9.11. 39. a = 147.74, 6 = 125.11, c = 198.53. 
40. ^ = 48° 26' 57", a = 73.229, 6 = 69.752, c = 91.724. 41. .4 = 
61° 17' 34", a = 105.4-5. 42. A = 78° 5& 20", B = 5S 3 13' 32", C= 
42° 51' S", a = 25.290, 6 = 21.909, c = 17,527. 



:anH.4 = 



(.Pii^ + P2P3 -TPzPi) (P1P2 —P%Pz +P3P1) 



etc., 



176 APPENDIX. 



a = PiPs - , etc. 43. c = 51.5, c = l 7 2 (a 2 + 6 2 — 2m 2 ), where m 
Pi sin J. 

is the medial line. 44. The formulae are : 

a = \V 2 (ra 2 2 + ra 3 2 ) — m x 2 , 6 = f ^ 2 (ra 3 2 + m?) — ra 2 2 , etc., 
where m 1? m 2 , m 3 are the medial lines to a, 6, c, respectively. 

Page 95. 

46. 45° 22 / 36" and 36° 54' 28". 47. 327.38, XF = CT cos H^~ -#) 

2 cos JjS cos JC 

48. radius = 10.689, chords = 20.508 and 18.121. 49.26.048. 50. 

XB = 2908.0 feet, ^4(7=3272.6 feet, .5(7=661.6 feet. 51. 1574.2. 

52. J.P= 775.9 feet, BJP = 622.1 feet, <7P = 225.1 feet. 

Page 96. 

54. 690.6. 57. B = 32.704, r = 13.304. If the angles are a and /?, 
and a is the distance between centres, 

R = a sin J (a -f- (3) cos J (a — /?), r = a cos J (a + /?) sin J (a — j3). 
58. 17°10 / 36". 

Page 106. 

1. 1.6953, 6.6170, 1765.5, 13.749. 2. 5.0051, 67.064, 5088.3, 32.742. 
3. x = 138° 3' 53", a = 6.4282. 4. = 69° 38' 49", p = .038706. 5. 
= 201° 38' 55", m = 116.37. 6. = 152° 52' 12", r = — 9.3442. 7. 
52° 26' 21" and 265° 51' 3". 8. —5° 35' 20" and 15° 58' 40". 9. 71° 
22' 21" and 153° 54' 11". 10. 306° 3' 48" and 48° 47' 56". 11. x = 
69° 49' 40", y = 58° 21' 56" ; x = 119° 0' 28", y = 9° IF 8" ; x = 92° 
42' 28", y = 81° 14' 44" ; x = 141° 53' 16", y = 32° 3' 56". 13. 1° 19' 
44" and 181° 19' 44". 14. —1° 49' 32" and 178° 10' 28". 15. 146° 
56' 59" and 29° 45' 1". 16. 19° 52' 21" and 102° 7 / 39". 17. 9° 54' 
56" and 189° 54' 56". 18. 71° 29' 15" and 152° 18' 45". 

19. tan y=zm, tan (%<j> + a;) = tan (45° — y) cot J0. 

20. tan y = m, tan (x — l<f) = tan (y — 45°) cot %$. 

21. tan 7 = m, sin (2a; — 0) = cot (7 -f 45°) sin 0. 

22. tan 7 = ra, cos (2a; — 0) = tan (7 + 45°) cos 0. 

23. 14° 0' 6" and 194° 0' 6". 

Page 107. 

24. 101° 55' 33" and 281° 55' 33". 25. 147° 15' 21" and 186° V 39". 
26. 110° 14' 56" and 207° 45' 3". 27. r = 24.268, = 119° 37' 34". 
28. m = 89.002, n = 30.066, x = 61° 9' 1", y = 87° 38' 20", or a; = 241° 
9' 1", ?/ = —92° 21' 40". 29. a; = 38° 28' 51", m = —91.184, or 
x = 218° 28' 51", m = 91.184. 30. = 303° 3' 13", = 38° 22' 44", 
r = 7.9228 . 31. = 176° 34' 51", 6 = 58° 36' 21", r = 8.0932. 



APPENDIX. 177 

Page 108. 

48. ^4 = 9° 55' 58", B = 19° 52' 55", c == 104.37. 49. a = 73.088, 
6 _. e = 178.70'. 50. a = 44.051, 6 = 66.789, e = 86.274. 51. 6 
= 499.74, c = 622.64, A = 15° 26' 14". 52. a = 329.41, 6 = 451.70, 
,B = 134° 56 / 10". 53. J. = 72° 15', J5 = 84° 48 / 45", 6 = 4.9795, 
c = 1.9486. 54. C= 32° 47 / 40", a = 97.555, 6 = 106.645, c = 58.300. 
55. B = 66° 28' 13", a = 21.953, b =25.169, c = 23.876. 56. B = 25° 
37' 54", a = 99.506, 6 = 43.873. 57. .4 = 61' 22' 30", a =107.39, 
6 = 83.42. 58. A = 40° 41' 33", B = 98° 15' 48", 6 = 494.80, c 
= 328.33. 59. B = 51° 53' 15", a = 82.147, 6 = 68.920. 60. A = 38° 
23' 50", i? = 73° 38' 5", 6 = 539.15. 61. A = 80° 36', a = 259.88, 
e = 247.53. 62. ,5 = 39° 36', 6 = 268.41, e = 81.31. 63. ^4=46° 
36 / 9", £ = 101° 31' 2", 6 = 56.834, e = 33.169. 64. B = 65° 24', 
a = 20.232, 6 = 18,512, c = 10.491. 65. # = 57° 23' 47", a = 122.11, 
6 = 104.46, e = 83.94. 66. A = 43° 5' 43", a = 17.763, c = 24.304. 
67. ^ = 30° 57', a = 15.884, 6 = 9.628, c = 18.717. 68. 6 = 46.171, 
A = 60° 30' 50", B = 78° 36' 5". 69. 45° 55'. 70. 40° 12'. 71. 78° 4'. 



Page 118. 

1. (1) cos 36° 4- i sin 36°, cos 108° 4- i sin 108°, —1, cos 252° + i sin 
252°, cos 324° 4- i sin 324° ; (2) cos 0° + * sin 0° = 1, cos 90° 4- i sin 
90° = i, cos 180° 4- i sin 180° = —1, cos 270° 4- i sin 270° = — i. 2. 3 
(cos 80° + i sin 80°), 3 (cos 200° + i sin 200°), 3 (cos 320° + i sin 320°). 



Page 125. 
1. 54° I'll". 2. 98° 20' 10". 

Page 126. 
1. 69°52 / 55". 2. 69° 44' 43". 



Page 148. 

18. sin \A = cos a . 19. Latitude, 30° 50' 14" N. ; Course, S. 
2 cos 2 ia 



cos a 
cos 2 j 
61° 31' 5" E. 20. 61°54 / 



Page 163. 



3. 3622.5 geographical miles. 
m 



178 APPENDIX 

Page 164. 

4. 32° 42' 45". 5. 81° 36 / 27", 71° 54' 24", 72° 58' 31", or their sup- 
plements. 6. b= 52° 46 / 40", ^1 = 50° 40' 10", C=83° 17 / 56". 7. 
^1 = 41° 2' 48", 6 = 113° V 42", c = 110° 55' 38". 

Page 170. 

1. (1) K= 1.9257 r\ r = 39° 28 / 12", R = 59° 57' 23" ; (2) iT = 
4.4383 r 2 , r = 63° 54 / 45", R = 84° 54' 38". 5. a = b = c = 109° 28 7 16". 



THE END. 



